if $w=x^2-2xy+3y^2,x=uv$ and $y= u^2-v^2$,use the chain rule to find $\frac{dw}{du}$ and $\frac{dw}{dv}$.
Here is what I did
\begin{alignat}{2}\frac{\partial w}{\partial u}&= \frac{\partial w}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial u} \\[1ex]
\frac{\partial w}{\partial x}&= 2x-2y &\hspace{-5em} \frac{\partial w}{\partial y}&= -2x+6y\\
\frac{\partial x}{\partial u}&= v&\hspace{-5em} \frac{\partial y}{\partial u}&= 2u \\[2ex]
\frac{\partial w}{\partial u}&= v(2x-2y)+2u(6y-2x)\\
&=2vx-2vy+12uy-4ux\\
&=2x(v-2u)+2y(6u-v)\\
&=2uv(v-2u)+2u^2-\rlap{2v^2(6u-v)}\\
&=(2uv^2-4u^2v)+\rlap{(12u^3-2u^2v-12uv^2+2v^3)}\\
&=-10uv^2-6u^2v+12u^3+2v^3\\
\end{alignat}
Am I on the right track to continue with $\dfrac{\partial w}{\partial v}$?