Suppose I have the following functions, $f(x,y)=xy-y^3$ and $g(u,v)=f(u^3,uv^2)$ then calculate the gradient $dg(u_0,v_0)$ using the chain rule.
My attempt: I set $x=u^3$ and $y=uv^2$
$\frac{\partial g}{\partial u}=\frac{\partial g}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial g}{\partial y}\frac{\partial y}{\partial u}=\frac{\partial g}{\partial x}(3u^2)+\frac{\partial g}{\partial y}(v^2)$
I think I understand it correctly up to this point, but how do I find $\frac{\partial g}{\partial x}$ and other terms like it?
As you set $x = u^3$, this implies $u = x^{1/3}$. Use this value of u in the expression above to find out the partial derivative of $3u^2$ w.r.t $x$.
I hope it will help you.