$$ x = x' \cos \theta - y'\sin \theta \\ y = x' \sin \theta + y'\cos \theta $$
$$ \frac{\partial f}{\partial x'} = \frac{\partial f}{\partial x} \cos \theta + \frac{\partial f}{\partial y} \sin \theta $$
are given. I don't understand how the partial derivative with respect to $x'$ is obtained. Can you explain?
$$ \frac{\partial^2 f}{\partial x'^2} = \frac{\partial^2 f}{\partial x^2}\cos^2\theta+ \frac{\partial }{\partial x}(\frac{\partial f}{\partial y})\sin\theta\cos\theta + \frac{\partial }{\partial y}(\frac{\partial f}{\partial x})\cos\theta\sin\theta + \frac{\partial^2 f}{\partial y^2}\sin^2\theta $$
You have $$ \begin{cases} x = x'\cos\theta - y'\sin\theta\\ y = x'\sin\theta + y'\sin\theta \end{cases} \quad\Rightarrow\quad \begin{cases} \frac{\partial x}{\partial x'} = \cos\theta\\ \frac{\partial y}{\partial x'} = \sin\theta\\ \end{cases} $$ so that you can apply the chain rule to get $$ \frac{\partial f}{\partial x'} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial x'} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial x'} = \frac{\partial f}{\partial x} \cos\theta + \frac{\partial f}{\partial y}\sin\theta. $$ Can you do the second partial derivative on your own now?