I'd like to know if my solution to this problem is good:
The change of variables $x = u + v, y = uv^2$ transforms $f (x,y)$ into $g(u, v)$. Compute the value of $\frac{\partial ^2g}{\partial v \partial u}$ at the point at which $u = 1, v = 1$, given that $\frac{\partial f}{\partial y}=\frac{\partial ^2f}{\partial x^2}=\frac{\partial ^2f}{\partial y^2}=\frac{\partial ^2f}{\partial x \partial y}=\frac{\partial ^2f}{\partial y \partial x}=1$, at that point.
My attempt to the solution:
$\frac{\partial g}{\partial u}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial u}=\frac{\partial f}{\partial x}\cdot1+1\cdot v^2$
$\frac{\partial ^2g}{\partial v \partial u}=\frac{\partial}{\partial v}(\frac{\partial g}{\partial u})=\frac{\partial}{\partial v}(\frac{\partial f} {\partial x}+v^2)=\frac{\partial}{\partial v}(\frac{\partial f}{\partial x})+2v=\frac{\partial}{\partial x}(\frac{\partial f}{\partial x})\cdot\frac{\partial x}{\partial v}+\frac{\partial}{\partial y}(\frac{\partial f}{\partial x})\cdot\frac{\partial y}{\partial v}+2v=\frac{\partial ^2f}{\partial x^2}+\frac{\partial ^2f}{\partial y \partial x}\cdot 2uv+2v$
if $u=1$ and $v=1$, and using the conditions given in the statement, $\frac{\partial ^2g}{\partial v \partial u}=(1)+(1)+2(1)(1)+2(1)=6$
If anyone could give me any feedback regarding my procedure I would greatly appreciate it, thanks.
In computing the first derivative for some reason you evaluated some, but not all, of the subexpressions at the point at which you’re supposed to be evaluating this second derivative, then went on to differentiate the resulting function. This evaluation was premature: you need to leave that until the end. Moreover, you were inconsistent about it: you evaluated $\partial f/\partial y$, in the resulting expression, but not $\partial f/\partial x$.
So, for the first derivative you should instead have $${\partial g\over\partial u} = {\partial f\over\partial x}{\partial x\over\partial u} + {\partial f\over\partial y}{\partial y\over\partial u} = {\partial f\over\partial x}+v^2{\partial f\over\partial y}.$$ Then, differentiate this with respect to $v$: \begin{align} {\partial\over\partial v}\left({\partial f\over\partial x}+v^2{\partial f\over\partial y}\right) &= {\partial\over\partial v}{\partial f\over\partial x}+{\partial\over\partial v}\left(v^2{\partial f\over\partial y} \right) \\ &= {\partial\over\partial v}{\partial f\over\partial x}+v^2{\partial\over\partial v}{\partial f\over\partial y}+\left({\partial\over\partial v}v^2\right){\partial f\over\partial y} \\ &= {\partial x\over\partial v}{\partial^2f\over\partial x^2} + {\partial y\over\partial v}{\partial^2f\over\partial x\partial y} + v^2\left({\partial x\over\partial v}{\partial^2f\over\partial x\partial y}+{\partial y\over\partial v}{\partial^2f\over\partial y^2}\right) + 2v{\partial f\over\partial y} \\ &= {\partial^2f\over\partial x^2} + 2uv{\partial^2f\over\partial x\partial y} + v^2\left({\partial^2f\over\partial x\partial y}+2uv{\partial^2f\over\partial y^2}\right) + 2v{\partial f\over\partial y}. \end{align} Having obtained this expression for ${\partial^2g\over\partial u\partial v}$ you can now substitute the numerical values of $u$, $v$ and the partial derivatives of $f$ into it.