Reasoning correct for continuity

Just for some background:

We defined continuity as follows: $f$ continuous in $x$ $\iff$ $\forall$ neigborhoods $V$ of $f(x)$, $\exists$ neighborhood $U$ of $x$, such that $f(U)\subseteq V$.

Problem at hand: Show that $f$ is continuous on $\mathbb R^{2}- \{0\}$

$f(x_{1},x_{2}) = \begin{cases} 0 & x=0\\ \frac{x_{1}x_{2}^{2}}{\vert\vert x\vert\vert_{2}^{4}} & x \neq 0 \\ \end{cases}$

I worked the solution out differently, however, my professor used the following succinct argument:

There is an $\epsilon > 0$ such that $B_{\epsilon}(x_{1},x_{2})\subseteq \mathbb R^{2}- \{0\}$, then it follows that $\forall (y_{1},y_{2})\in B_{\epsilon}(x_{1},x_{2}): f(y_{1},y_{2})=\frac{y_{1}y_{2}^{2}}{\vert\vert y\vert\vert_{2}^{4}}$. From this he follows that $f$ is continuous on $B_{\epsilon}(x_{1},x_{2})$.

Question: Is this basically just a narrowed down version of the definition I mentioned at the top that I will prove below? If I were to write it down according to the definition, I'd would do the following:

Let $V$ be any neighborhood of $f(x)$, by definition $\exists \delta > 0$, such that $B_{\delta}(f(x_{1},x_{2}))\subseteq V$ then we are free to choose an $\epsilon > 0$ such that $f(B_{\epsilon}(x_{1},x_{2}))\subseteq B_{\delta}(f(x_{1},x_{2}))\subseteq V$ q.e.d (Is this correct?)

Question 1: I have only proven continuity in $x$, however, my professor has proven the continuity on $B_{\epsilon}(x_{1},x_{2})$, but I don't understand how he has done it if he is using the definition above?

Question 2: Is it trivial that $B_{d}(f(x_{1},x_{2}))=f(B_{d}(x_{1},x_{2}))$ or not? Any counterexamples?

Your professor is choosing a neighborhood of $x=(x_1,x_2)$ which does not include $0$, so you can use
$$\forall (y_{1},y_{2})\in B_{\epsilon}(x_{1},x_{2}): f(y_{1},y_{2})=\frac{y_{1}y_{2}^{2}}{\vert\vert y\vert\vert_{2}^{4}}$$
• $B_{d}(f(x_{1},x_{2}))=f(B_{d}(x_{1},x_{2}))$ is not true. The RHS is not either an open ball. – Mohammad Riazi-Kermani Jun 14 '18 at 21:52