Partial Derivatives using ChainRule Can any one please explain the second step:-
Step1:
$$\frac{\partial }{\partial x}\left[(1-x^2)\frac{\partial u}{\partial x}\right]+\frac{\partial }{\partial y}\left[y^2\frac{\partial u}{\partial y}\right]=0$$
Step2:
$$L.H.S.=-2x\frac{\partial u}{\partial x}+(1-x^2)\frac{\partial ^2u}{\partial x^2}+2y\frac{\partial u}{\partial y}+y^2\frac{\partial ^2u}{\partial y^2}$$

Are you sure about the +1 after step 2? I think, that 1 is disappearing after derivation.
If you derive $-x^2\frac{\partial U}{\partial x}$ then you have to use the product rule.
$$u=-x^2$$
$$v=\frac{\partial U}{\partial x}$$
$$(u \cdot v)'=u'\cdot v+u\cdot v'$$
