It's not the fine way to present an operator product of two noncommuting operators in a seemeingly symmetric way  $$x^{-1} p^2 \ne p^2 x^{-1}.$$ The operator is not symmetric let alone selfadjoint. Take the first one and try to find a measure space such that

$$ \int\ \rho(x) \ f^*(x) x^{-1} g''(x) dx = -\int \ \rho(x) \ f''^*(x) x^{-1} g(x) dx $$

Background: In order to define an eigenvalue problem $A f - \lambda f= 0$ in an infinite dimensional space of functions, unbounded operators like $x, \partial_x$ have to be restricted to a domain in Hilbert space, that yields finite scalar products $$\left< f, A g \right> = \int f(x) A g(x) dx \lt \infty.$$ In oder to get real eigenvalues $A$ has to be symmetric: $$\left< f, A g \right> = \int f(x)^* A g(x) dx = \int (A f(x))^* g(x) dx .$$ If an operator of derivatives has left coefficients depending on x, symmetry by integration by parts shifts the coefficents from left to right $$\int f(x) a(x) \partial_{xx} g(x) dx = -\int \partial_x (f(x) a(x)) \partial_x g(x) dx = \int \partial_{xx}\left(( a(x) f(x)\right) g(x) dx$$

There is only one way to get rid of the unsymmetric product: consider another Hilbert space with integration measure $a(x)^{-1} dx$ cancelling the coefficient.

This is a common problem, if the Laplacian is expressed in a curvilinear coordinates. Eg polar

 $$\partial_{xx}+\partial_{yy} = \frac{1}{r}\partial_r r \partial_r + \frac{1}{r^2} \partial_{\phi\phi}.$$

The radial factor of the Hilbert space is transformed into

$$ \left< f(r) , g(r) \right> = \int_0^\infty \rho(r)\ f(r) \ g(r) dr$$ such that, in the scalar product of derivatives, the factor r disapears

$$\int_0^\infty f(r) (r^{-1}\partial_r (r g'(r))) r dr = -\int_0^\infty f'(r) \ r \ g'(r) dr = \int_0^\infty g(r) (r^{-1}\partial_r (r f'(r))) r dr$$

The expression in the middle has the required symmetry for the linear space of of functions with integrable derivatives diverging not faster than $1/\sqrt r$ at $r=0$.

It's not a fine way to present an operator product of two noncommuting operators in a seemingly symmetric way, as $$x^{-1} p^2 \ne p^2 x^{-1}.$$ Your operator is not symmetric, let alone self-adjoint. 

Take the first one and try to find a measure space such that $$ \int\ \rho(x) \ f^*(x) x^{-1} g''(x) dx = -\int \ \rho(x) \ f''^*(x) x^{-1} g(x) dx .$$

Background: In order to define an eigenvalue problem, $A f - \lambda f= 0$, in an infinite-dimensional space of functions, unbounded operators like $x, \partial_x$ have to be restricted to a domain in Hilbert space that yields finite scalar products $$\left< f, A g \right> = \int f(x) A g(x) dx \lt \infty.$$ In oder to get real eigenvalues, $A$ has to be symmetric: $$\left< f, A g \right> = \int f(x)^* A g(x) dx = \int (A f(x))^* g(x) dx .$$ If an operator of derivatives has left coefficients depending on x, the symmetry, by integration by parts, shifts the coefficients from left to right, $$\int f(x) a(x) \partial_{xx} g(x) dx = -\int \partial_x (f(x) a(x)) \partial_x g(x) dx = \int \partial_{xx}\left(( a(x) f(x)\right) g(x) dx.$$

There is only one way to get rid of the unsymmetric piece product: to consider another Hilbert space with integration measure $a(x)^{-1} dx$ cancelling that coefficient.

This is a common problem, if the Laplacian is expressed in a curvilinear coordinates. E.g., polar  $$\partial_{xx}+\partial_{yy} = \frac{1}{r}\partial_r r \partial_r + \frac{1}{r^2} \partial_{\phi\phi}.$$

The radial factor of the Hilbert space transforms to $$ \left< f(r) , g(r) \right> = \int_0^\infty \rho(r)\ f(r) \ g(r) dr$$ such that, in the scalar product of derivatives, the factor r disapears $$\int_0^\infty f(r) (r^{-1}\partial_r (r g'(r))) r dr = -\int_0^\infty f'(r) \ r \ g'(r) dr = \int_0^\infty g(r) (r^{-1}\partial_r (r f'(r))) r dr.$$ The expression in the middle has the required symmetry for the linear space of of functions with integrable derivatives diverging not faster than $1/\sqrt r$ at $r=0$.

It's not the fine way to present an operator product of two noncommuting operators in a seemeingly symmetric way $$x^{-1} p^2 \ne p^2 x^{-1}.$$ The operator is not symmetric let alone selfadjoint. Take the first one and try to find a measure space such that

$$ \int\ \rho(x) \ f^*(x) x^{-1} g''(x) dx = -\int \ \rho(x) \ f''^*(x) x^{-1} g(x) dx $$

It's not the fine way to present an operator product of two noncommuting operators in a seemeingly symmetric way $$x^{-1} p^2 \ne p^2 x^{-1}.$$ The operator is not symmetric let alone selfadjoint. Take the first one and try to find a measure space such that

$$ \int\ \rho(x) \ f^*(x) x^{-1} g''(x) dx = -\int \ \rho(x) \ f''^*(x) x^{-1} g(x) dx $$

Background: In order to define an eigenvalue problem $A f - \lambda f= 0$ in an infinite dimensional space of functions, unbounded operators like $x, \partial_x$ have to be restricted to a domain in Hilbert space, that yields finite scalar products $$\left< f, A g \right> = \int f(x) A g(x) dx \lt \infty.$$ In oder to get real eigenvalues $A$ has to be symmetric: $$\left< f, A g \right> = \int f(x)^* A g(x) dx = \int (A f(x))^* g(x) dx .$$ If an operator of derivatives has left coefficients depending on x, symmetry by integration by parts shifts the coefficents from left to right $$\int f(x) a(x) \partial_{xx} g(x) dx = -\int \partial_x (f(x) a(x)) \partial_x g(x) dx = \int \partial_{xx}\left(( a(x) f(x)\right) g(x) dx$$

There is only one way to get rid of the unsymmetric product: consider another Hilbert space with integration measure $a(x)^{-1} dx$ cancelling the coefficient.

This is a common problem, if the Laplacian is expressed in a curvilinear coordinates. Eg polar

$$\partial_{xx}+\partial_{yy} = \frac{1}{r}\partial_r r \partial_r + \frac{1}{r^2} \partial_{\phi\phi}.$$

The radial factor of the Hilbert space is transformed into

$$ \left< f(r) , g(r) \right> = \int_0^\infty \rho(r)\ f(r) \ g(r) dr$$ such that, in the scalar product of derivatives, the factor r disapears

$$\int_0^\infty f(r) (r^{-1}\partial_r (r g'(r))) r dr = -\int_0^\infty f'(r) \ r \ g'(r) dr = \int_0^\infty g(r) (r^{-1}\partial_r (r f'(r))) r dr$$

The expression in the middle has the required symmetry for the linear space of of functions with integrable derivatives diverging not faster than $1/\sqrt r$ at $r=0$.