I am stuck in understanding how to proceed to show if the following functions admit partial derivatives at those specific points. We have been taught not to use the definition but rather to, say, evaluate $f(x, 0)$ and $f(0, y)$, that is along the axis, and then take their derivatives and evaluating them at the critical points.
Yet for these functions I’m stuck
$$f_1(x, y) = \begin{cases} \frac{x^4+y^4}{xy} & x \neq 0, y \neq 0 \\\\ 0 & \text{else} \end{cases}$$
$$f_2(x, y) = \begin{cases} \frac{x^2-y^2-1}{1-x} & x^2-y^2 \neq 1, x \neq 1 \\\\ K & \text{else} \end{cases}$$
So for what concerns the first function, I am stuck because if I evaluate $f(x, 0)$ I don’t understand if this doesn’t exist, or if this is $0$ (according to the definition of $f$…?). But if it were $0$ then the derivative would exist at any $(0, y)$ and vice versa for $(x, 0)$.
BUT on the other side, $f(x, x) = 2x^2$ whose derivative is $4x$ which is zero at any $(0, y)$
For the second function I have the same doubts: I found that $K$ should be $-2$ for the continuity, but in fact even if $K = -2$ there is no continuity. Now, $f(1, y)$ doesn’t exist or is it $0$?
Also, say $f(x, 0) = -(x+1)$ hence the derivative does exist at $(1, 0)$?
I never faced similar exercises, I ask for some help. Thank you!
Since $f_1$ is constant on the $x$-axis, its partial derivative with respect to $x$ at any point of the form $(x,0)$ is $0$. On the other hand if $x\ne0$, its partial derivative with respect to $y$ at $(x,0)$ does not exist since at such a point, $f_1$ is not even continuous with respect to $y$.
By symmetry, this answers your questions for $f_1$ at every point of the two axes (including the point $(0,0)$, where the two partial derivatives are $0$).
For the same reasons, the partial derivative of $f_2$ with respect to $y$ at any point of the form $(1,y)$ is $0$, and its partial derivative with respect to $x$ at $(1,y)$ can (perhaps) exist only if $f_2$ is continuous w.r.t. $x$ at that point, i.e. if $y=0$ and $K=-2$. Finally, if $K=-2$, $\frac{\partial f_2}{\partial x}(1,0)=\lim_{x\to1}\frac{\frac{x^2-1}{1-x}+2}{x-1}$ actually exists, and equals $-1$.