Solution (a) The gradient of $f$ at $(x,y)$ is $${\bf\nabla}\! f(x,y)=f_{x}(x,y)\mathbf{i}+f_{y}(x,y)\mathbf{j}=3x^{2}y \mathbf{i}+x^{3}\mathbf{j}$$

The gradient of $f$ at $(2,1)$ is $${\bf\nabla}\! f(2,1)=12\mathbf{i}+8\mathbf{j}$$

(b) The unit vector $\mathbf{u}$ from $(2,1)$ to $(3,5)$ is $$\begin{equation*} \mathbf{u=}\dfrac{( 3-2) \mathbf{i} +( 5-1) \mathbf{j} }{\sqrt{( 3-2) ^{2}+( 5-1) ^{2}}}=\frac{\mathbf{i}+4 \mathbf{j}}{\sqrt{17}} \end{equation*}$$

We use formula (3) for the directional derivative to find $D_{\mathbf{u} }(2,1).$ $$\begin{equation*} D_{\mathbf{u}}(2,1)={\bf\nabla\! }f(2,1)\,{\bf\cdot}\, \mathbf{u}=(12\mathbf{i}+8 \mathbf{j})\,{\bf\cdot}\, \frac{\mathbf{i}+4\mathbf{j}}{\sqrt{17}}=\frac{44}{\sqrt{17} } \end{equation*}$$

The directional derivative of $f$ at $(2,1)$ in the direction of $\mathbf{u}$ is $\dfrac{44}{\sqrt{17}}$