Let's start with an example:
Determine the value of the positive integer $p$ for which the function
$$ f(x,y) = \begin{cases} \frac{(x-y)^{p}}{x^2+y^2} & \text{if $(x,y) \neq (0,0)$} \\ 0 & \text{if $(x,y)= (0,0)$} \end{cases} $$ is differentiable at $(0,0)$
I understand that the definition of differentiability of multivariable function begins with the notion of the equation of a tangent plane (or hyperplane):
$$h(a,b) = f(a,b) + f_x(a,b)(x-a)+f_y(a,b)(y-b)$$
And the function is said to be differentiable at $(0,0)$ if $$\lim_{(x,y)\to (0,0)} \frac{f(x,y)-h(x,y)}{\sqrt{x^2+y^2}} = 0$$ holds. But for the question in the example, I do not even know if the partial derivative exists. I then try to assume that it exists and evaluate the equation of the hyperplane by expressing the partial derivatives in terms of $x, y$ and then producing the equation of the tangent plane. Then:
$$h(x,y) = \frac{(x-y)^{p}(p-2)}{x^2+y^2}$$
And if $(x,y)=(0,0)$, the equation seems to be not well-defined. In the case of not even having a candidate tangent plane to start with, how should I chosse $p$ to make the function differentiable at $(0,0)$? And more importantly, what are some general methods for this type of question?
We have that
$$f_x(0,0)=\lim_{x\to 0}\frac{x^p}{x^2}=\lim_{x\to 0} x^{p-2}$$
and
$$f_y(0,0)=\lim_{y\to 0}\frac{y^p}{y^2}=\lim_{y\to 0} y^{p-2}.$$
If $p=1$ the partial derivatives don't exist; if $p=2$ we have $f_x(0,0)=f_(0,0)=1$ and if $p\ge 3$ then $f_x(0,0)=f_y(0,0)=0.$
Assume $p=2:$
$$\lim_{(x,y)\to (0,0)} \frac{f(x,y)-h(x,y)}{\sqrt{x^2+y^2}} = \lim_{(x,y)\to (0,0)} \dfrac{\dfrac{(x-y)^2}{x^2+y^2}-x-y}{\sqrt{x^2+y^2}}.$$ But
$$\lim_{x\to 0} \frac{f(x,0)-h(x,0)}{\sqrt{x^2}} = \lim_{x\to 0} \dfrac{\dfrac{x^2}{x^2}-x}{\sqrt{x^2}}$$ doesn't exist. So $f$ is not differentible at $(0,0).$
Assume $p\ge 3:$
$$\lim_{(x,y)\to (0,0)} \frac{f(x,y)-h(x,y)}{\sqrt{x^2+y^2}} = \lim_{(x,y)\to (0,0)} \dfrac{\dfrac{(x-y)^p}{x^2+y^2}}{\sqrt{x^2+y^2}}=\lim_{(x,y)\to (0,0)} \dfrac{(x-y)^p}{(x^2+y^2)^{3/2}}.$$
If $p=3$ the limit doesn't exist (Consider the limit along the paths $(x,kx)$). So $f$ is not differentible at $(0,0).$
If $p\ge 4$ the limit is $0.$ So $f$ is differentible at $(0,0).$