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

We're given the following function :

$$f(x,y)=\dfrac{1}{1+x-y}$$

Now , how to prove that the given function is differentiable at $(0,0)$ ?

I found out the partial derivatives as $f_x(0,0)=(-1)$ and $f_y(0,0)=1$ ,

Clearly the partial derivatives are continuous , but that doesn't guarantee differentiability , does it ?

Is there any other way to prove the same ?

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.