Let $V$ be a finite-dimensional Vector space over a Field $K$ and $v,w \in V\setminus\{0\}$. Let $\lambda$ be an eigenvalue of $f \in \operatorname{End}_K(V)$ and $\mu$ be an eigenvalue of $g \in \operatorname{End}_K(V)$.
Show that $\lambda\mu$ is an eigenvalue of $f \otimes g \in \operatorname{End}_K(V \otimes_K V)$.
Since $f,g$ are endomorphisms, there exists exactly one endomorphism $f \otimes g: V \otimes_K V \to V \otimes_K V$ such that $(f \otimes g)(v \otimes w) = f(v) \otimes g(w)$. Then it follows that $f(v) \otimes g(w) = \lambda v \otimes \mu w = \lambda\mu(v \otimes w)$. This means that $\lambda\mu$ is an eigenvalue of $f \otimes g$.
I'm not really sure if this is correct or not, so any correction or verification is greatly appreciated.
Yes, your answer is correct. That is, $\lambda \mu$ is an eigenvalue of $f \otimes g$ with associated eiegnvector $v \otimes w$.
Note that there is no need to say "there exists exactly one endomorphism $f \otimes g: V \otimes_K V \to V \otimes_K V$ such that $(f \otimes g)(v \otimes w) = f(v) \otimes g(w)$" since this is part of the definition of $f \otimes g$. If you choose to keep this in, then you should clarify that $f \otimes g$ is defined such that $(f \otimes g)(x \otimes y) = f(x) \otimes g(y)$ holds for all vectors $x \in V$ and $y \in W$, not just the given eigenvectors $x = v$ and $y = w$.