$z = e^y \phi \left(y e^ \frac{x^2}{2y^2} \right)$ satisfies $(x^2 - y^2) \frac{\partial z}{\partial x} + xy \frac{\partial z}{\partial y} = xyz$ Show that the function $z = e^y \phi \left(y e^ \frac{x^2}{2y^2}\right)$ satisfies the equation $(x^2 - y^2) \frac{\partial z}{\partial x} + xy \frac{\partial z}{\partial y} = xyz$.
$\frac{\partial z}{\partial x} = \dfrac{x}{y} e^ {y + \frac{x^2}{2y^2}} \phi ' \left(y e^ \frac{x^2}{2y^2}\right) $. In the same way $\frac{\partial z}{\partial y}$ is calculated. And then I did the multiplications but the calculation becomes very long.
Is there another way, for example with a change of variable, so that it is not so long?

$$
\frac{dz}{dx} = e^y \phi'\frac{x}{y}e^{\frac{x^2}{2y^2}}
$$
$$
\frac{dz}{dy} = e^y\phi + e^y\phi'\left(e^\frac{x^2}{2y^2} - \frac{x^2}{y^2}e^{\frac{x^2}{2y^2}}\right)
$$
$$
(x^2-y^2)\frac{dz}{dx} + xy\frac{dz}{dy} = e^y \phi'\frac{x}{y}e^{\frac{x^2}{2y^2}}(x^2-y^2)+\left(e^y\phi + e^y\phi'\left(e^\frac{x^2}{2y^2} - \frac{x^2}{y^2}e^{\frac{x^2}{2y^2}}\right)\right)xy
$$
$$
e^y\phi xy+e^y\phi'\left(\frac{x}{y}e^{\frac{x^2}{2y^2}}(x^2-y^2)+\left(e^\frac{x^2}{2y^2} - \frac{x^2}{y^2}e^{\frac{x^2}{2y^2}}\right)xy\right)
$$
$$
e^y\phi xy+e^y\phi'\left(\frac{x^3}{y}e^{\frac{x^2}{2y^2}}-xye^{\frac{x^2}{2y^2}}+xye^\frac{x^2}{2y^2} - \frac{x^3}{y}e^{\frac{x^2}{2y^2}}\right) = xy e^y\phi = xyz
$$
$$
\therefore (x^2-y^2)\frac{dz}{dx} + xy\frac{dz}{dy} = xyz \quad z = e^y \phi
$$
