Show that this two variable limit don't exist Prove that the following limit doesn't exist: $$\lim_{(x,y)\to(1,0)}\frac{\sin(1-x^2+y^2)-y}{xy}$$ 
I guess I need to find different values for different directions.
 A: Consider $x=\cosh t$ and $y=\sinh t$:
$$
\lim_{t\to0}\frac{\sin(1-\cosh^2t+\sinh^2t)-\sinh t}{\sinh t\cosh t}=
\lim_{t\to0}-\frac{1}{\cosh t}=-1
$$
because $\cosh^2t-\sinh^2t=1$.
Now try $x=1+t$, $y=t$:

$$\lim\limits_{t\to0}\dfrac{\sin(1-1-2t-t^2+t^2)-t}{t(1+t)}=\lim\limits_{t\to0}\dfrac{\sin(-2t)-t}{t(1+t)}=\lim\limits_{t\to0}\dfrac{-2\cos2t-1}{1+2t}=-3$$

A: First direction: $y=1-x,$ second direction:  $y=-(1-x).$
If $y=1-x$ then 
$$\begin{gather}\lim_{\substack{{(x,y)\to(1,0)}\\{y=1-x}}}\frac{\sin(1-x^2+y^2)-y}{xy}=
\lim_{x\to{1}}\frac{\sin((1-x)(1+x)+(1-x)^2)-(1-x)}{x(1-x)}=\\
=\lim_{x\to{1}}\frac{\sin{\left((1-x)(1+x+1-x)\right)-(1-x)}}{x(1-x)}=
\lim_{x\to{1}}\frac{\sin{\left(2(1-x)\right)-(1-x)}}{x(1-x)} =1\end{gather}$$
since $\lim\limits_{t\to{0}}\dfrac{\sin{t}}{t}=1.$
But for  $y=-1+x$ we have
$$\begin{gather}\lim_{\substack{{(x,y)\to(1,0)}\\{y=-1+x}}}\frac{\sin(1-x^2+y^2)-y}{xy}=
\lim_{x\to{1}}\frac{\sin((1-x)(1+x)+(1-x)^2)+(1-x)}{x(x-1)}=\\
=\lim_{x\to{1}}\frac{\sin{\left((1-x)(1+x+1-x)\right)+(1-x)}}{x(x-1)}=\\
=\lim_{x\to{1}}\frac{\sin{\left(2(1-x)\right)+(1-x)}}{x(x-1)} =-3.\end{gather}$$
