My attempt:
Given $$(x^{2}-y^{2})\frac{\partial u}{\partial x}+2xy\frac{\partial u}{\partial y} =0$$ By the method of characteristic, we have $$\frac{dx}{x^{2}-y^{2}} =\frac{dy}{2xy}=dt\Rightarrow\int2xydx=\int(x^{2}-y^{2})dy\Rightarrow x^{2}y=x^{2}y-\frac{1}{3}y^{3}+c\Rightarrow y^{3}=c'$$ for some constant $c'$.Hence, the general solution is $u=f(y^{3})$ for some differentiable function $f$.
But I was told that this is incorrect. I'm having trouble understanding where I messed up.
The method of characteristic reduces a PDE problem into an ODE problem, in this case to the ODE $\frac{dx}{x^2-y^2} = \frac{dy}{2xy}$ as you identified. Writing this in the form $2xy\ dx + (y^2-x^2)\ dy = 0$, we notice that this is homogenous and so can be solved by substituting $x = uy, dx = udy + ydu$ to get $$2uy^2(udy + ydu) + y^2(1-u^2)dy = 0 \implies 2uy^3du + y^2(1+u^2)dy,$$ giving us the separable equation $$\frac{2u}{1+u^2}du + \frac{1}{y}dy = 0$$ which can be solved by integrating the terms independently. The strategy you used did not work because the equation was not separated, so integration is not an effective solution method.