Differentiability of a two variable function $f(x,y)=\dfrac{1}{1+x-y}$

You have several ways to approach the question.

First one $f$ is the ratio of two differentiable functions, the denominator one not vanishing in the neighborhood of the origin. Hence $f$ is differentiable at the origin.

Second one Using a theorem stating that if $f$ is continuous in an open set $U$ and has continuous partial derivatives in $U$ then $f$ is continuously differentiable at all points in $U$.

Third one Using the definition of the derivative, prove that $$\lim\limits_{(h,k) \to (0,0)} \frac{f(h,k)-f(0,0)+h-k}{\sqrt{h^2+k^2}}=0$$

If all partial derivatives of a function (over all possible variables) are continuous at some point, then the function is differentiable at that point.

You need to show that the limit definition of the partial derivative from each direction is the same value. Hence

$\frac{\partial f}{\partial x}(0,0) = \lim_{h \to 0^{-}} \frac{f(0+h,0)-f(0,0)}{h} = \lim_{h \to 0^{-}} \frac{\frac{1}{1+h} - 1}{h} = \lim_{h \to 0^{-}} -\frac{1}{(1+h)^{2}} = -1$ (Using l'hopital's rule).

$\frac{\partial f}{\partial x}(0,0) = \lim_{h \to 0^{+}} \frac{f(0+h,0)-f(0,0)}{h} = \lim_{h \to 0^{-}} \frac{\frac{1}{1+h} - 1}{h} = \lim_{h \to 0^{-}} -\frac{1}{(1+h)^{2}} = -1$.

Therefore the function is differentiable with respect to $x$. Now rinse and repeat the same thing for $y$ to determine if $f$ is differentiable with respect to $y$.