[tex]\begin{bmatrix}u\\v\end{bmatrix}=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}x\cos\theta-y\sin\theta\\x\sin\theta+y\cos\theta\end{bmatrix}[/tex]
so that
[tex]\begin{cases}u(x,y)=x\cos\theta-y\sin\theta\\v(x,y)=x\sin\theta+y\cos\theta\end{cases}[/tex]
For a function [tex]f(u,v)=f(u(x,y),v(x,y))[/tex], we have by the chain rule,
[tex]\dfrac{\partial f}{\partial x}=\dfrac{\partial f}{\partial u}\dfrac{\partial u}{\partial x}+\dfrac{\partial f}{\partial v}\dfrac{\partial v}{\partial x}[/tex]
and
[tex]\begin{cases}\frac{\partial u}{\partial x}=\cos\theta\\\\\frac{\partial v}{\partial x}=\sin\theta\end{cases}[/tex]
[tex]\implies\dfrac{\partial f}{\partial x}=\cos\theta\dfrac{\partial f}{\partial u}+\sin\theta\dfrac{\partial f}{\partial v}[/tex]
Let [tex]g(u,v)=\frac{\partial f}{\partial u}[/tex] and [tex]h(u,v)=\frac{\partial f}{\partial v}[/tex]. This substitution is made just to make the application of the chain rule clearer.
[tex]\dfrac{\partial f}{\partial x}=\cos\theta\,g+\sin\theta\,h[/tex]
Differentiating again wrt [tex]x[/tex] gives
[tex]\dfrac{\partial^2f}{\partial x^2}=\cos\theta\dfrac{\partial g}{\partial x}+\sin\theta\dfrac{\partial h}{\partial x}[/tex]
By the chain rule,
[tex]\dfrac{\partial g}{\partial x}=\dfrac{\partial g}{\partial u}\dfrac{\partial u}{\partial x}+\dfrac{\partial g}{\partial v}\dfrac{\partial v}{\partial x}[/tex]
and our substitution shows that, for instance,
[tex]\dfrac{\partial g}{\partial u}=\dfrac{\partial}{\partial u}\dfrac{\partial f}{\partial u}=\dfrac{\partial^2f}{\partial u^2}[/tex]
and so
[tex]\dfrac{\partial g}{\partial x}=\cos\theta\dfrac{\partial^2f}{\partial u^2}+\sin\theta\dfrac{\partial^2f}{\partial v\partial u}[/tex]
Similarly, we find
[tex]\dfrac{\partial h}{\partial x}=\cos\theta\dfrac{\partial^2f}{\partial u\partial v}+\sin\theta\dfrac{\partial^2f}{\partial v^2}[/tex]
Putting everything together, we have
[tex]\dfrac{\partial^2f}{\partial x^2}=\cos^2\theta\dfrac{\partial^2f}{\partial u^2}+\cos\theta\sin\theta\dfrac{\partial^2f}{\partial v\partial u}+\sin\theta\cos\theta\dfrac{\partial^2f}{\partial u\partial v}+\sin^2\theta\dfrac{\partial^2f}{\partial v^2}[/tex]
and we can similarly find that
[tex]\dfrac{\partial^2f}{\partial y^2}=\sin^2\theta\dfrac{\partial^2f}{\partial u^2}-\cos\theta\sin\theta\dfrac{\partial^2f}{\partial v\partial u}-\sin\theta\cos\theta\dfrac{\partial^2f}{\partial u\partial v}+\cos^2\theta\dfrac{\partial^2f}{\partial v^2}[/tex]
Adding together these derivatives, we see the mixed partials cancel, and recalling that [tex]\cos^2\theta+\sin^2\theta=1[/tex], we end up with
[tex]\dfrac{\partial^2f}{\partial x^2}+\dfrac{\partial^2f}{\partial y^2}=\dfrac{\partial^2f}{\partial u^2}+\dfrac{\partial^2f}{\partial v^2}[/tex]
as required.
