Find all points where $f$ is continuous $f:\mathbb{R}^2 \rightarrow \mathbb{R}$
$f(x,y)=\begin{cases}
\frac{\sin{y}+2yx^2}{y},  & y \gt0 \\
\frac{y+3x}{1-y}, & y\le 0
\end{cases}$
I have to check
a) What points is $f$ continuous at?
b) Is $f$ class $C^1$ at $(1,0)$
My idea for a)
Function $f$ is continuous at all points $(x_0,y_0)$ where $y_0 \neq 0$ as sum, difference and quotient of continuous functions. Now for $y_0=0$, I have to check if $\lim_{x \to x_0} f(x,0) = f(x_0,0)$. I don't know what to plug in instead of $f(x,0)$ in limit.
b) I can find differential $Df(1,0)$ and check if it is continuous in $(1,0)$? I am not sure about b)
 A: As you say, the function is continuous for all $y \neq 0$, and so we need to consider $y=0$.
$$\lim_{y \to 0^+} \left(\frac{\sin y +2x^2y}{y}\right) \ = \ \lim_{y\to 0^+}\left(\frac{\sin y}{y} + \frac{2x^2y}{y}\right) \ = \ 1+\lim_{y \to 0^+}\left(\frac{2x^2y}{y}\right) \ = \ 1+2x^2$$
$$\lim_{y\to 0^-}\left(\frac{3x+y}{1-y}\right) \ = \ 3x$$
For the function to be continuous, we need the $x$- and $y$-limits to agree.
We need (at least) $1+2x^2=3x$, i.e. $2x^2-3x+1=0$, i.e. $x=\frac{1}{2}$ or $x=1$.
The function is discontinuous at the the points $y=0$ and $x \neq \frac{1}{2},1$.
We need to consider $x=\frac{1}{2}$ and $x=1$ separately. 
$$f(\tfrac{1}{2},y) = \left\{ \begin{array}{ccc} \frac{2\sin y + y}{2y} & : & y > 0 \\ \frac{2y+3}{2-2y} & : & y \le 0 \end{array}\right.$$
As $y \to 0^-$, $f \to \frac{3}{2}$. As $y \to 0^+$, $f \to \frac{3}{2}$ and so $f(\frac{1}{2},y)$ is continuous at $y=0$.
$$f(1,y) = \left\{ \begin{array}{ccc} \frac{\sin y + 2y}{y} & : & y > 0 \\ \frac{3+y}{1-y} & : & y \le 0 \end{array}\right.$$
As $y \to 0^-$, $f \to 3$. As $y \to 0^+$, $f \to 3$. It follows that $f$ is continuous at $(x,y)=(1,0)$.
Although the function is continuous as $(1,0)$, it needs to be continuous in a neighbourhood of $(1,0)$ for it to be differentiable. The function is not continuous at $(1+\varepsilon,0)$ for arbitrarily small $\varepsilon>0$.
