Verify function solves heat equation $\Delta f = \partial_t f$ So the notorious function is written as $f(x,t) = t^{-n/2}\,e^{\frac{-x^2}{4\,t}} $
It strikes me as that I have to show that the first derivative with respect to t equals the sum of both second derivatives. But on the other side this seems irritating because the first derivative looks already complicated:
$$\partial_tf(x,t) = -\frac{n}{2} \,t^{-n/2-1} \,e^{\frac{-x^2}{4\,t}}+ t^{-n/2}\,\frac{x^2}{4\,t^2}\, e^{\frac{-x^2}{4\,t}}.$$ Should this really equal $\Delta f = \partial_x^2f +\partial_t^2 f$?
I'm not really confident whether this might be the task or is it?

It is a little annoying, but that's just how it goes sometimes. First of all, since you have it written as $x^2,$ you're clearly working with $n=1.$ Otherwise, this should be $\|x\|^2.$
Calculate that
$$\partial_x f=-\frac{1}{2}t^{-3/2} xe^{-x^2/4t},$$ and $$\partial_{xx} f=-\frac{1}{2}t^{-3/2} e^{-x^2/4t}+\frac{1}{4}t^{-5/2}x^2e^{-x^2/4t}.$$ When $n=1,$ this is $\Delta f$ and is exactly the same as what you computed for $\partial_t f$ (modulo a sign error on the power of $t$ in the first term: it should read $t^{-n/2-1}$).
