Differentiability at a vertical tangent point

For $x > 3$, $f(x) - f(3) = (x-3)^{1/2} - 0$, so $\frac{f(x)-f(3)}{x-3} = \frac{(x-3)^{1/2}}{x-3} = (x-3)^{-1/2}$, which $\to +\infty$ as $x \to 3^+$. For $x < 3$, $f(x) - f(3) = (3-x)^{1/2}$, so $\frac{f(x)-f(3)}{x-3} = \frac{(3-x)^{1/2}}{-(3-x)} = -(3-x)^{-1/2}$, which $\to -\infty$ as $x \to 3^-$.

Since the two one-sided limits are $+\infty$ and $-\infty$ (both infinite, and not equal), the limit defining $f'(3)$ does not exist, so $f$ is not differentiable at $x=3$. Graphically, this is because the curve has a cusp at $x=3$, where the one-sided derivatives have opposite infinite values.