The function $f:\mathbb R\to\mathbb R$ defined by $f(x)=\mathbf 1_{x\gt1}$ is not Gâteaux-differentiable at $x=1$ since $t^{-1}(f(x+th)-f(x))=t^{-1}$ diverges when $t\to0$, $t\gt0$, for $x=1$ and $h=1$. In fact, in dimension $1$, Gâteaux-differentiability implies continuity.
On the other hand, consider a point $z$ in $\mathbb R^2$, a circle $\bar C$ passing through $z$, $C=\bar C\setminus\{z\}$, and the function $g=\mathbf 1_C$.
For every $h\ne(0,0)$ in $\mathbb R^2$, the line $z+h\mathbb R$ does not meet $C$ in a neighborhood of $z$, that is, there exists some positive $t_{h,z}$ such that $\{z+th\mid|t|\lt t_{h,z}\}\cap C$ is empty. (To see this, note that $C$ intersects each line passing through $z$ at one point at most, which is never $z$.)
In particular, for every $|t|\lt t_{h,z}$, $t^{-1}(g(z+th)-g(z))=0$ hence the Gâteaux-differential of $g$ at $z$ in the direction $h$ exists and is $0$, although $g$ is not continuous at $z$ since the point $z$ is in the boundary of $C$.
No: A function which is differentiable at $x$ is continuous at $x$. To prove this, note that the quantity
$$\left|\frac{f(x) - f(y)}{x - y}\right|$$
is a bounded function of $y$ in a deleted neighborhood containing $x$; if $M$ is a bound, then rearranging shows that
$$|f(x) - f(y)| \le M |x - y|$$
for all $y$ in the deleted neighborhood (and in fact, in the neighborhood).
Best Answer
To begin with we take $x_0 = 0$ : we want $f(0) = 0$ and $f'$ defined at $0$, with $f'(0)=1$.
Then you take :
$f(x) = x$ if $x \in \mathbb{Q}$
$f(x) = x + x^2$ if $x \notin \mathbb{Q}$
$f$ is continuous only at $0$. Moreover, you can check that $f'(0) = 1$ : indeed we have $f(0) = 0$, so $\mid \frac{f(x) - f(0)}{x} - 1\mid = \mid \frac{f(x)}{x} - 1\mid$. This quantity equals either to $0$ if $x \in \mathbb{Q}$ or to $\mid x \mid$ if $x \notin \mathbb{Q}$. Either way, when $x$ approaches $0$, $\mid \frac{f(x) - f(0)}{x} -1 \mid $ approaches $0$. Although $f$ is almost nowhere continuous, the derivative at $0$ exists and $f'(0) = 1$.
More generally, if you want $g(x_0) = y_0$, $g'(x_0) = a$ and $g$ continuous only at $x_0$, you can take :
$g(x) = y_0 + a\times f(x-x_0)$
A quite equivalent (but slightly different) explicit version would be :
$g(x) = y_0 + a(x-x_0)$ if $x\in \mathbb{Q}$
$g(x) = y_0 + a(x-x_0) +(x-x_0)^2$ if $x\notin \mathbb{Q}$