This is what I did, but I'm not sure if this is right.
First I find the second-order partial derivatives, by using the chain rule:
$\frac{\partial u}{\partial \theta}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial \theta}\Rightarrow\frac{\partial u}{\partial x}\frac{\partial x}{\partial \theta}=\frac{\partial u}{\partial \theta}-\frac{\partial u}{\partial y}\frac{\partial y}{\partial \theta}\Rightarrow\frac{\partial u}{\partial x}=\frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial x}-\frac{\partial u}{\partial y}\frac{\partial y}{\partial \theta}\frac{\partial \theta}{\partial x}$
$\frac{\partial u}{\partial x}=-\frac{\partial u}{\partial \theta}\cdot\frac{y}{x^2+y^2}+\frac{\partial u}{\partial y}r\cos(\theta)\cdot\frac{y}{x^2+y^2}$
$\frac{\partial u}{\partial r}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial r}\Rightarrow\frac{\partial u}{\partial y}\frac{\partial y}{\partial r}=\frac{\partial u}{\partial r}-\frac{\partial u}{\partial x}\frac{\partial x}{\partial r}\Rightarrow\frac{\partial u}{\partial y}=\frac{\partial u}{\partial r}\frac{\partial r}{\partial y}-\frac{\partial u}{\partial x}\frac{\partial x}{\partial r}\frac{\partial r}{\partial y}$
$\frac{\partial u}{\partial y}=\frac{\partial u}{\partial r}\cdot\frac{y}{(x^2+y^2)^\frac{1}{2}}-\frac{\partial u}{\partial x}\cos(\theta)\cdot\frac{y}{(x^2+y^2)^\frac{1}{2}}$
$\frac{\partial^2 u}{\partial y \partial x}=\frac{\partial}{\partial y}(\frac{\partial u}{\partial x})=-\frac{\partial u}{\partial \theta}\frac{\partial}{\partial y}(\frac{y}{x^2+y^2})+\frac{\partial u}{\partial y}r \cos \theta\frac{\partial}{\partial y}(\frac{y}{x^2+y^2})$
$\frac{\partial^2 u}{\partial y \partial x}=-\frac{\partial u}{\partial \theta}\cdot\frac{x^2-y^2}{(x^2+y^2)^2}+\frac{\partial u}{\partial y}r\cos(\theta)\cdot\frac{x^2-y^2}{(x^2+y^2)^2}$
$\frac{\partial^2 u}{\partial y \partial x}=-\frac{\partial u}{\partial \theta}\cdot\frac{r^2cos(2\theta)}{r^4}+\frac{\partial u}{\partial y}r\cos(\theta)\cdot\frac{r^2cos(2\theta)}{r^4}$
$\frac{\partial^2 u}{\partial y^2}=\frac{\partial u}{\partial r}\frac{\partial}{\partial y}(y\cdot(x^2+y^2)^{-\frac{1}{2}})-\frac{\partial u}{\partial x}\cos(\theta)\frac{\partial}{\partial y}(y\cdot(x^2+y^2)^{-\frac{1}{2}})$
$\frac{\partial^2 u}{\partial y^2}=\frac{\partial u}{\partial r}(\frac{1}{\sqrt{x^2+y^2}}-\frac{3}{2}2y^2(x^2+y^2)^{-\frac{3}{2}})-\frac{\partial u}{\partial x}\cos(\theta)(\frac{1}{\sqrt{x^2+y^2}}-\frac{3}{2}2y^2(x^2+y^2)^{-\frac{3}{2}})$
$\frac{\partial^2 u}{\partial y^2}=\frac{\partial u}{\partial r}(\frac{1}{r}-\frac{3}{2}2r^2\sin^2(\theta)r^\frac{1}{2})-\frac{\partial u}{\partial x}\cos(\theta)(\frac{1}{r}-\frac{3}{2}2r^2\sin^2(\theta)r^\frac{1}{2})$
Now using the fact that $x=r \cos(\theta)$ and $y=r \sin(\theta)$, I find $\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial y}$ in terms of $r$ and $\theta$:
$\frac{\partial u}{\partial r}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial r}\Rightarrow\frac{\partial u}{\partial x}\cos (\theta)+\frac{\partial u}{\partial y}\sin(\theta)$
$\frac{\partial u}{\partial \theta}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial \theta}\Rightarrow\frac{\partial u}{\partial x}\left(-r\right)\sin(\theta)+\frac{\partial u}{\partial y}r \cos(\theta)$
Writing in terms of a matrix:
$\begin{bmatrix}
\frac{\partial u}{\partial r}\\
\frac{\partial u}{\partial \theta}
\end{bmatrix}=\begin{bmatrix}
\cos \theta & \sin \theta\\
-r \sin \theta & r \cos \theta
\end{bmatrix}\begin{bmatrix}
\frac{\partial u}{\partial x}\\
\frac{\partial u}{\partial y}
\end{bmatrix}$
$\begin{bmatrix}
\frac{\partial u}{\partial x}\\
\frac{\partial u}{\partial y}
\end{bmatrix}=\begin{bmatrix}
\cos \theta & \sin \theta\\
-r \sin \theta & r \cos \theta
\end{bmatrix}^{-1}\begin{bmatrix}
\frac{\partial u}{\partial r}\\
\frac{\partial u}{\partial \theta}
\end{bmatrix}$
$\begin{bmatrix}
\frac{\partial u}{\partial x}\\
\frac{\partial u}{\partial y}
\end{bmatrix}=\begin{bmatrix}
\cos \theta & -\frac{\sin \theta}{r}\\
\sin \theta & \frac{\cos \theta}{r}
\end{bmatrix}\begin{bmatrix}
\frac{\partial u}{\partial r}\\
\frac{\partial u}{\partial \theta}
\end{bmatrix}$
From here it's just a simple case of plugging in terms into the expression, which I am too lazy to do. Can anyone confirm that this is indeed correct?
You're on the right track. I would view things as follows.
Let $Q=f(q,p)$ and $p=g(q,Q)$. Then, we have both
$$dQ=f_1(q,p)dq+f_2(q,p)dp \tag 1$$
and
$$dp=g_1(q,Q)dq+g_2(q,Q)dQ \tag 2$$
Substitution of $dQ$ as given in $(1)$ into $(2)$ yields
$$\begin{align}
dp&=g_1(q,Q)dq+g_2(q,Q)\left(f_1(q,p)dq+f_2(q,p)dp \right)\\\\
&=\left(g_1(q,Q)+g_2(q,Q)f_1(q,p)\right)dq +f_2(q,p)g_2(q,Q)\,dp\tag 3
\end{align}$$
Inasmuch as the differential terms on both sides of $(3)$ must be equal, we obtain
$$\bbox[5px,border:2px solid #C0A000]{f_2(q,p)g_2(q,Q)=\frac{\partial Q}{\partial p}\times \frac{\partial p}{\partial Q}=1}$$
which was to be shown along with
$$g_1(q,Q)+g_2(q,Q)f_1(q,p)=\frac{\partial p}{\partial q}+\frac{\partial p}{\partial Q}\times \frac{\partial Q}{\partial q}=0$$
Best Answer
Your second derivative is wrong:
$$ w_u=\frac{u}{\sqrt{u^2+v^2}} \Rightarrow w_{uu}=\frac{\sqrt{u^2+v^2}-u\frac{u}{\sqrt{u^2+v^2}}}{u^2+v^2}=\frac{u^2+v^2-u^2}{(u^2+v^2)\sqrt{u^2+v^2}} $$