We have two functions :
$1) \;f(x,y)=\begin{cases} \frac{xy}{x^2-y^2}&\text{when }\;|x|\neq|y| \\0&\text{when }\;|x|=|y| \end{cases}$
$2) \;g(x,y)=\begin{cases} x^2\sin\frac1x+y^2&\text{when }\;(x,y)\in \{\mathbb{R} \setminus{0}\} \times \mathbb{R} \\y^2&\text{when }\;(x,y)\in\{0\}\times\mathbb{R} \end{cases}$
I want to check if they are differentiable in $(0,0)$
My work so far :
$1)$ Let's take sequences :
$x_n=\frac1n,\;y_n=\frac2n$. Then $ f(x_n,y_n)=\dfrac{\frac1n\cdot \frac2n}{\frac{1}{n^2}-\frac{4}{n^2}}=-\dfrac{2}{3}\nrightarrow0.$ So it's not continous at $(0,0)$ so it cannot be differentiable in that point.
$2)$ Let's take sequences :
$x_n=\frac{1}{\frac{\pi}{2}-\pi n},\;y_n=0$. Then $$g(x_n,y_n)=(\frac{1}{\frac{\pi}{2}-\pi n})^2 \cdot \sin(\frac{\pi}{2}-\pi n)=(\frac{1}{\frac{\pi}{2}-\pi n})^2 \cdot \cos(n\pi)=(\frac{1}{\frac{\pi}{2}-\pi n})^2 \cdot (-1)^n.$$
And the above sequence doesn't has limit to this function is not continuous at (0,0), so it cannot be continuous.
Am I thinking correctly ?
For the last limit, what you did is wrong. $g(x_n,y_n)$ has a limit. Because $(\frac{1}{\frac{\pi}{2}-\pi n})^2$ converges to $0$ and $(-1)^n$ is bounded...