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 $r=0$.