The solution said the function $f$ is not differentiable at $0$. For if $g: \mathbb{R}^2 \rightarrow \mathbb{R}$ is a function that is differentiable at $o$, then $Dg(0)$ is a $1$ by $2$ matrix of the form [a b], and $g'(0; u) = ah + bk$, which is a linear function of $u$.
I don't understand how the book got $g'(0; u) = ah + bk$. Anyone could help, thanks ahead.
It's saying that if the jacobian of this map, $[f_x \quad f_y]=[a \quad b]$, is a linear map $\mathbb{R}^2\to\mathbb{R}$. If $u\in\mathbb{R}^2$ is a vector, say $u=\begin{bmatrix}h \\ k\end{bmatrix}$, then the derivative of $g$ at zero in the direction of $u$ is (I am guessing) $$g'(0;u)=[a \quad b]\begin{bmatrix}h \\ k\end{bmatrix}=ah+bk$$
Remember the formula $$df = \frac {\partial f}{\partial x}dx +\frac {\partial f}{\partial y}dy.$$
In your question $h$ is the $x$ increment and $k$ is the $y$ increment and $a$ is partial derive of $f$ with respect to $x$ at $(0,0)$ and $b$ is the partial derivative of $f$ with respect to $y$ at (0,0).
This is just the formula $$df = \frac {\partial f}{\partial x}dx +\frac {\partial f}{\partial y}dy.$$ with different notation.
