= Solution
Work on an open domain with $xy>0$ and $x\ne0$, where the displayed square root is real and differentiable. Split $u=v+h$ with $v=\phi(xy)$ and $h=\sqrt{xy}\,\psi(y/x)$. For the first term, the <chain rule> gives
$$
v_{xx}=y^2\phi''(xy),\qquad v_{yy}=x^2\phi''(xy),
$$
so $x^2v_{xx}-y^2v_{yy}=0$.
For the second term, set $t=y/x$. Its first <partial derivatives> are
$$
h_x=\frac{\sqrt{xy}}x\left(\frac12\psi(t)-t\psi'(t)\right),\qquad
h_y=\frac{\sqrt{xy}}y\left(\frac12\psi(t)+t\psi'(t)\right).
$$
They show $xh_x+yh_y=h$, the degree-one case of <Euler theorem for homogeneous functions>. Differentiating that identity with respect to each variable yields
$$
xh_{xx}+yh_{xy}=0,\qquad xh_{xy}+yh_{yy}=0.
$$
The mixed <partial derivatives> agree here by the <chain rule>. Multiplying these equations by $x$ and $y$, respectively, proves the <second derivative identity for a degree-one homogeneous function>:
$$
x^2h_{xx}=-xyh_{xy}=y^2h_{yy}.
$$
Adding the results for $v$ and $h$ verifies the <partial differential equation>
$$
\boxed{x^2u_{xx}-y^2u_{yy}=0.}
$$
Equivalently, $h(\lambda x,\lambda y)=\lambda h(x,y)$ for positive $\lambda$, so the same proof follows directly from its <homogeneous function> property. The statement is local to a domain where the given expression is defined; it does not require assigning values on $x=0$.
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