In the Physics.SE, I have seen this very nice question where the acceleration $\mathbf{a}$ is obtained using the differential operator $\boldsymbol \nabla$.
$$\mathbf{a}=\frac{d\mathbf{v}}{dt}=\color{red}{\frac{dx}{dt}\frac{\partial\mathbf{v}}{\partial x}+\frac{dy}{dt}\frac{\partial\mathbf{v}}{\partial y}+\frac{dz}{dt}\frac{\partial\mathbf{v}}{\partial z}=(\mathbf{v}\cdot\boldsymbol \nabla)\:\mathbf{v}}$$
Why can I obtained the expression coloured in red?
(A particular case of) the multivariable chain rule says: $$\frac{d}{dt}f\big(x_1(t),\ldots,x_n(t)\big)=\sum_{i=1}^n\frac{\partial f}{\partial x_i}\frac{dx_i}{dt}.$$ In your specific case: $$\frac{d}{dt}\mathbf{v}\big(x(t),y(t),z(t)\big)=\frac{\partial\mathbf{v}}{\partial x}\frac{dx}{dt}+\frac{\partial\mathbf{v}}{\partial y}\frac{dy}{dt}+\frac{\partial\mathbf{v}}{\partial z}\frac{dz}{dt}$$ Now the result follows from noting that $(\mathbf{v}\cdot\nabla)\mathbf{v}$ is equal to: $$(\mathbf{v}\cdot\nabla)\mathbf{v}=\left(\frac{dx}{dt}\frac{\partial}{\partial x}+\frac{dy}{dt}\frac{\partial}{\partial y}+\frac{dz}{dt}\frac{\partial}{\partial z}\right)\mathbf{v}.$$