Let $u=\ln x$, $v = \ln y$. Then what is another form of $$x^2 \left( \frac{\partial^2 f}{\partial x^2}\right) + y^2 \left(\frac{\partial^2 f}{\partial y^2}\right) + x \left( \frac{\partial f}{\partial x}\right) + y \left(\frac{\partial f}{\partial y}\right)=0 $$ in terms only of the second-order partial derivatives $\frac{\partial^2 f}{\partial u^2} $ and $\frac{\partial^2 f}{\partial v^2}$?
Attempt. I solved for $u$ and $v$, that is $x = e^u$ and $y=e^v$. Then, after substitution, I obtained $$e^{2u} \left( \frac{\partial^2 f}{\partial u^2}\right) + e^{2v} \left(\frac{\partial^2 f}{\partial v^2}\right) + e^u \left( \frac{\partial f}{\partial u}\right) + e^v \left(\frac{\partial f}{\partial v}\right)=0 $$
What could be the next for this?
I don't think that your calculations are entirely correct. It seems like you've made a direct replacement of $df/dx$ with $df/du$ and $d^2f/dx^2$ with $d^2f/du^2$ and so on. You can't do this because there is a dependence between $u$ and $x$ (and between $v$ and $y$) that you have to account for. Using the chain rule,
$$\frac{df}{du} = \frac{df}{dx}\frac{dx}{du} = \frac{df}{dx}e^u = x\frac{df}{dx}$$
$$\frac{d^2f}{du^2} = \frac{d}{du}\left(\frac{df}{du}\right) = \frac{d}{dx}\left(x\frac{df}{dx}\right)\frac{dx}{du} = \left(\frac{dx}{dx}\frac{df}{dx} +x\frac{d^2f}{dx^2}\right)e^u = x\frac{df}{dx}+x^2\frac{d^2f}{dx^2}$$
Similarily, you can conclude that
$$\frac{df}{dv} = y\frac{df}{dy}$$
$$\frac{d^2f}{dv^2} = y\frac{df}{dy}+y^2\frac{d^2f}{dy^2}$$
Inserting this into the original equation yields
$$\frac{d^2f}{du^2} + \frac{d^2f}{dv^2} = 0$$