Upside-Down Triangle, Squared: What the?

First we evaluate the second partial derivative of the function with respect to $x$ . Then we evaluate the second partial derivative of the function with respect to $y$ . Executing we find that

$\frac{\partial^2 f}{\partial x^2} =-ye^{ix}-\frac{y}{x^2}+2$

and

$\frac{\partial^2 f}{\partial y^2}=0$

Once we have found our partial derivatives, we then find the linear combination of the two - per the definition of the Laplacian of a scalar function. Now we have that

$\nabla^2f(x, \; y)= \frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}= 2- \left (ye^{ix}+\frac{y}{x^2} \right )$

and finally, using Euler's Identity $e^{ix}=i \sin x +\cos x$ , we find the answer to the question:

$\nabla^2 f \left (\frac{\pi}{2}, \; 3 \right )=2- \left ( 3i \sin\left ( \frac{\pi}{2} \right ) +3\cos\left ( \frac{\pi}{2} \right )+\frac{12}{\pi^2}\right )=\boxed{2-\left ( 3i+\frac{12}{\pi^2} \right )}$