There are no examples in the text to go by, so I am not sure which direction to go with this one:
Let $f(x,y) = \frac{x^4 + y^4}{x^2 + y^2}$ if $x^2 + y^2 \neq 0$, and define $f(0,0)=0$. Show that $f$ has first partial derivatives at all points, satisfying the inequalities $|f_1(x,y)|\leq 6|x|$, $|f_2(x,y)| \leq 6|y|$. Is $f$ differentiable at $(0,0)$?
**Note, $f_1$ and $f_2$ are the partials with respect to $x$ and $y$ respectively. $$f_1 = \frac{4x^3(x^2 + y^2) - 2x(x^4 + y^4)}{(x^2 + y^2)^2}$$ $$f_2 = \frac{4y^3(x^2 + y^2) - 2y(x^4 + y^4)}{(x^2 + y^2)^2}$$
Except, when $f(0,0) = 0$, then $$f_1 = 0$$ $$f_2 = 0$$
I do not understand how to make the connection with each of these being less than or equal to $6|x|$ and $6|y|$ respectively.
An easier way to see if $f$ is differentiable at $(0,0)$: We know $f_1(0,0) = 0,f_2(0,0)=0.$ That tells you $Df(0,0),$ if it exists, is just the $0$ transformation. So hint: Is it true that
$$f(x,y) - f(0,0) = 0 + o(\sqrt {x^2+y^2})$$
as $(x,y) \to (0,0)?$