In the solutions, my professor wrote that the limit is equal to zero and then proved it using the squeeze theorem. However, he didn't explain why the limit is zero.
I remember an exercise from calculus I (this is calculus II) which was similar but for one variable, and my then professor explained that when you have a limited function like sine or cosine multiplied by something that tends to... I think it was zero, the limit is zero.
Is this why the limit tends to zero and does it apply in this case? If yes, why is it true?
According to the definition of limit, if we assume that $f(x,y)\longrightarrow 0$ as $(x,y)\longrightarrow(0,0)$ and $|g(x,y)|\leq M$, for every $\varepsilon/M > 0$ there exists $\delta > 0$ such that \begin{align*} 0 < \|(x,y)\| < \delta \Longrightarrow |f(x,y)| < \frac{\varepsilon}{M} \Longrightarrow |f(x,y)g(x,y)| < \frac{\varepsilon|g(x,y)|}{M} \leq \frac{\varepsilon M}{M} = \epsilon \end{align*} from whence we conclude that $f(x,y)g(x,y)\longrightarrow 0$ as $(x,y)\longrightarrow(0,0)$.
At your case, $f(x,y) = x^{2} + y^{2}$, $\displaystyle g(x,y) = \sin\left(\frac{1}{xy}\right)$ and $M = 1$.
Hopefully this helps.