I feel like I have done everything correctly, but I end up on a dead end. Question:
\begin{equation}
    \label{simple_equation0}
 u = {\varphi }(xy)+\sqrt{xy}{\psi}\left(\frac{y}{x}\right)
\end{equation}
Show that 
\begin{equation}
    \label{simple_equation1}
 x^{2}\frac{\partial^2 u}{\partial x^{2}}-y^{2}\frac{\partial^2 u}{\partial y^{2}}= 0
\end{equation}
This is my work so far Where did I go wrong?
 A: We will need $\phi$ and $\psi$ to be twice differentiable, and we note beforehand that we better restrict the domain to something like  $x,y>0$ so that the fractions and square roots involved behave well. We will use this subsequently without mention, e.g., by equating expressions like $\frac1x\sqrt {xy}$ and $\sqrt{\frac yx}$.

By linearity, we can consider the $\phi$-summand and the $\psi$-summand separately.
First compute the partial derivatives of $\let\phi\varphi(x,y)\mapsto \phi(xy)$.
By the chain rule, 
$$\frac\partial{\partial x}\phi(xy)=y\phi'(xy) $$
and then 
$$\frac{\partial^2}{\partial x^2}\phi(xy)=\frac\partial{\partial x}y\phi'(xy)=y^2\phi''(xy). $$
By symmetry, likewise
$$\frac{\partial^2}{\partial y^2}\phi(xy)=x^2\phi''(xy)$$
and so indeed 
$$x^2\frac{\partial^2}{\partial x^2}\phi(xy)- y^2\frac{\partial^2}{\partial y^2}\phi(xy)=0.$$
We treat the $\psi$-summand accordingly, it merely involves a bit more complicated derivatives:
By product and chain rule,
$$\begin{align}\frac\partial{\partial y}\bigl(\sqrt{xy}\psi(\tfrac yx)\bigr)&=\sqrt{xy}\frac\partial{\partial y}\psi(\tfrac yx)+\psi(\tfrac yx)\frac\partial{\partial y}\sqrt{xy}\\&=
\sqrt{\frac yx}\psi'(\tfrac yx)+\frac12\psi(\tfrac yx)\sqrt{\frac xy}\end{align}$$
and 
$$\begin{align}\frac\partial{\partial x}\bigl(\sqrt{xy}\psi(\tfrac yx)\bigr)&=\sqrt{xy}\frac\partial{\partial x}\psi(\tfrac yx)+\psi(\tfrac yx)\frac\partial{\partial x}\sqrt{xy}\\&=
-yx^{-2}\sqrt{\frac yx}\psi'(\tfrac yx)+\frac12\psi(\tfrac yx)\sqrt{\frac yx}\\
&=-\left(\frac yx\right)^{\frac 32}\psi'(\tfrac yx)+\frac12\psi(\tfrac yx)\sqrt{\frac yx}.\end{align}$$
So it does get a bit more complicate dthis time. I urge you to check and continue this pre-coffee (hence error-prone) calculation for the second derivative. I won't. 
Instead allow we to show how one could come up with the form of solution in the first place. For this, we consider only simple functions, say power series, or even simpler polynomials, so ultimately a monomial $(x,y)\mapsto x^ny^m$. 
What happens to this under the operator $x^2\frac{\partial^2}{\partial x^2}-y^2\frac{\partial^2}{\partial y^2}$? Fortunately, this is much easier to answer as it clearly produces
$$ \bigl(n(n-1) - m(m-1)\bigr)x^ny^m.$$
This is $\equiv 0$ iff $n(n-1)=m(m-1)$. The obvious solutions are $m=n$ (leading to $(xy)^n$ as in the $\phi$-summand) and $m-1=-n$, or a bit more symmetrically $m-\frac12=-(n-\frac12)$. This leads to   $x^ny^m=\sqrt{xy}\left(\frac yx\right)^{m-\frac12}$ as in the $\psi$-summand.
A: $\def\fr{\left(\frac yx\right)}$
We will assume $x>0,y>0$ everywhere.
$$
\frac{\partial u}{\partial x}=y\phi'(xy)+\frac12\frac{y^{1/2}}{x^{1/2}}\psi\fr-\frac{y^{3/2}}{x^{3/2}}\psi'\fr
$$
$$
\frac{\partial u}{\partial y}=x\phi'(xy)+\frac12\frac{x^{1/2}}{y^{1/2}}\psi\fr+
\frac{y^{1/2}}{x^{1/2}}\psi'\fr
$$
Differentiating again we obtain:
$$\begin{align}
\frac{\partial^2 u}{\partial x^2}&
=y^2\phi''(xy)-\frac14\frac{y^{1/2}}{x^{3/2}}\psi\fr
+\frac{y^{3/2}}{x^{5/2}}\psi'\fr 
+\frac{y^{5/2}}{x^{7/2}}\psi''\fr\\
\frac{\partial^2 u}{\partial y^2}&
=x^2\phi''(xy)-\frac14\frac{x^{1/2}}{y^{3/2}}\psi\fr
+\frac{1}{x^{1/2}y^{1/2}}\psi'\fr 
+\frac{y^{1/2}}{x^{3/2}}\psi''\fr
\end{align}$$
Can you take it from here?
