Suppose that $u=u(x,y)$ and $v=v(x,y)$ have continuous second partial derivatives. If for each $f$, we have $$\frac{\partial^2f}{\partial u^2}+\frac{\partial^2f}{\partial v^2}=\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2},$$ prove that the Jacobian $\begin{pmatrix}\frac{\partial u}{\partial x}&\frac{\partial u}{\partial y}\\\frac{\partial v}{\partial x}&\frac{\partial v}{\partial y}\end{pmatrix}$ is constant.
By using the chain rule, $$\frac{\partial^2f}{\partial x^2}=\left(\frac{\partial u}{\partial x}\right)^2\frac{\partial^2f}{\partial u^2}+2\frac{\partial u}{\partial x}\frac{\partial v}{\partial x}+\left(\frac{\partial v}{\partial x}\right)^2\frac{\partial^2f}{\partial v^2}\\ \frac{\partial^2f}{\partial y^2}=\left(\frac{\partial u}{\partial y}\right)^2\frac{\partial^2f}{\partial u^2}+2\frac{\partial u}{\partial y}\frac{\partial v}{\partial y}+\left(\frac{\partial v}{\partial y}\right)^2\frac{\partial^2f}{\partial v^2}$$ and we see that the Jacobian is orthogonal. But how to prove that it is constant?
Bad calculation keeps me away from solving this.
As is pointed out by @peabody, we have $$\frac{\partial^2f}{\partial x^2}=\left(\frac{\partial u}{\partial x}\right)^2\frac{\partial^2f}{\partial u^2}+2\frac{\partial u}{\partial x}\frac{\partial v}{\partial x}+\left(\frac{\partial v}{\partial x}\right)^2\frac{\partial^2f}{\partial v^2}+\frac{\partial^2u}{\partial x^2}\frac{\partial f}{\partial u}+\frac{\partial^2v}{\partial x^2}\frac{\partial f}{\partial v}\\ \frac{\partial^2f}{\partial y^2}=\left(\frac{\partial u}{\partial y}\right)^2\frac{\partial^2f}{\partial u^2}+2\frac{\partial u}{\partial y}\frac{\partial v}{\partial y}+\left(\frac{\partial v}{\partial y}\right)^2\frac{\partial^2f}{\partial v^2}+\frac{\partial^2u}{\partial y^2}\frac{\partial f}{\partial u}+\frac{\partial^2v}{\partial y^2}\frac{\partial f}{\partial v}$$ rather than I wrote before.
Using the given condition (since $f$ is arbitrary) we get $$ \frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}=0\\ \left(\frac{\partial u}{\partial x}\right)^2+\left(\frac{\partial u}{\partial y}\right)^2=1 $$
From the second identity $$ \frac{\partial^2u}{\partial x^2}\frac{\partial u}{\partial x}\frac{\partial u}{\partial y}+\frac{\partial^2u}{\partial x\partial y}\left(\frac{\partial u}{\partial y}\right)^2=0\\ \frac{\partial^2u}{\partial x\partial y}\left(\frac{\partial u}{\partial x}\right)^2+\frac{\partial^2u}{\partial y^2}\frac{\partial u}{\partial x}\frac{\partial u}{\partial y}=0 $$
Adding up we get $$\frac{\partial^2u}{\partial x\partial y}=0.$$
Let $\frac{\partial u}{\partial x}=A(x)$, $\frac{\partial u}{\partial y}=B(y)$. Put this in $\left(\frac{\partial u}{\partial x}\right)^2+\left(\frac{\partial u}{\partial y}\right)^2=1$ we get that both are constant.
Similarly $\frac{\partial v}{\partial x}$ and $\frac{\partial v}{\partial y}$ are constant. Hence $\begin{pmatrix}\frac{\partial u}{\partial x}&\frac{\partial u}{\partial y}\\\frac{\partial v}{\partial x}&\frac{\partial v}{\partial y}\end{pmatrix}$ is constant.