Let \(I\subseteq\RR\) be an open interval, let \(f\colon I\to\RR\), and let \(x_0\in I\). If \(f\) is differentiable at \(x_0\), then \(f\) is continuous at \(x_0\).
For \(x\in I\) with \(x\neq x_0\), \[ f(x)-f(x_0)=(x-x_0)\,\frac{f(x)-f(x_0)}{x-x_0}. \] As \(x\to x_0\), the first factor tends to \(0\) and the second tends to \(f'(x_0)\) by differentiability. Hence \(f(x)-f(x_0)\to0\), so \(f\) is continuous at \(x_0\).
The converse fails. The function \(f\colon\RR\to\RR\), \(f(x)=\abs{x}\), is continuous at \(0\). Its difference quotient \(\frac{f(h)-f(0)}{h}=\frac{\abs{h}}{h}\) equals \(1\) for \(h>0\) and \(-1\) for \(h<0\), so it has no limit as \(h\to0\), and \(f\) is not differentiable at \(0\).