Partial Legendre transform: understanding a simple example Consider the following function:
$$f(x_1, x_2) = x_1^2x_2+x_1x_2^3$$
$f$ is a function of $(x_1, x_2)$. The conjugate variables $(u_1, u_2)$ to $(x_1, x_2)$ are $$u_1 = \partial f/ \partial x_1 = 2 x_1 x_2 + x_2^3$$ and $$u_2 = \partial f/ \partial x_2 = x_1^2 + 3x_1x_2^2.$$
One can construct $$g=f - u_1 x_1$$ which, to my understanding, replaces $x_1$ to its conjugate variable, $u_1$. The differential of $g$ is 
$$dg = u_2 dx_2 - x_1 du_1$$
thus $g=g(u_1, x_2)$. In words, $g$ is a function of the old variable $x_2$ and the conjucate variable to $x_1$, $u_1$. 
Now I fail to see this in the above example. I can compute 
$$g = f - u_1x_1 = -x_1^2x_2$$
but $g$ is then still a function of $(x_1, x_2)$, not $(u_1, x_2)$. What am I missing here? 
In theory, I could invert the $u_1=u_1(x_1, x_2)$ and $u_2=u_2(x_1, x_2)$ equations, then plug $x_1(u_1, u_2)$ and $x_2(u_1, u_2)$ into $f(x_1, x_2)$ to get $f(u_1, u_2)$. But that's just a simple change of variable. What's the point of the Legendre transformation concept then? 
Also, in this particular example, inverting the $u(x)$ equations doesn't seem obvious. How would one get the expression for $g(u_1, x_2)$?
 A: I can show you the steps that you need to calculate the Legendre transformation, you have to look at the theory behind .
for a given function of two variable $f(x,y)$ we are looking for the Legendre function $\tilde{f}(u,v)$
Step I:
The determinant of matrix $A$ must be unequal zero.  where $A$ is:
$$A= \left[ \begin {array}{cc} {\frac {\partial ^{2}}{\partial {x}^{2}}}f
 \left( x,y \right) & \left( {\frac {\partial }{\partial x}}f \left( x
,y \right)  \right) {\frac {\partial }{\partial y}}f \left( x,y
 \right) \\  \left( {\frac {\partial }{\partial x}}f
 \left( x,y \right)  \right) {\frac {\partial }{\partial y}}f \left( x
,y \right) &{\frac {\partial ^{2}}{\partial {x}^{2}}}f \left( x,y
 \right) \end {array} \right]
$$
Step II:
with the equations:
$$u=\frac{\partial f(x,y)}{\partial x}\tag 1$$
$$v=\frac{\partial f(x,y)}{\partial y}\tag 2$$
we get from equation (1) , $x=f_x(u,y)$ and from equation (2)  $y=f_y(u,x)$
both solutions must be exist
Step III:
with this solutions $f_x$ and $f_y$ we obtain the Legendre function
$$\tilde{f}(u,v)=u\,f_x(u,v)+v\,f_y(u,v)-f(f_x(u,v),f_y(u,v))$$
Example:
$$f(x,y)={x}^{2}y+x{y}^{3}$$
$$f_x(u,y)=1/2\,{\frac {u-{y}^{3}}{y}}$$
$$f_y(u,x)=1/3\,{\frac {\sqrt {3}\sqrt { \left( x \right)  \left( u-{x}^{2}
 \right) }}{x}}
$$
$$\tilde{f}(u,v)=1/2\,{\frac {u \left( u-{v}^{3} \right) }{v}}+1/3\,\sqrt {3}\sqrt {v
 \left( u-{v}^{2} \right) }-1/12\,{\frac { \left( u-{v}^{3} \right) ^{
2}\sqrt {3}\sqrt {v \left( u-{v}^{2} \right) }}{{v}^{3}}}-1/18\,{
\frac { \left( u-{v}^{3} \right) \sqrt {3} \left( v \left( u-{v}^{2}
 \right)  \right) ^{3/2}}{{v}^{4}}}
$$
