Solution (a) We begin by finding the gradient of $f$ at $(1,-2)$. $$\begin{array}{rcl@{\qquad}l} {\bf\nabla\! }f(x,y) &=&(2x-y)\mathbf{i}+(2y-x)\mathbf{j} & \color{#0066A7}{{\bf\nabla}\! f(x,y)=f_{x}(x,y) \mathbf{i}+f_{y}(x,y)\mathbf{j}} \\ {\bf\nabla\! }f(1,-2) &=&4\mathbf{i}-5\mathbf{j} & \color{#0066A7}{{x=1; y=-2}} \end{array}$$

The direction $\mathbf{u}$ for which $D_{\mathbf{u}}f(1,-2)$ is maximum occurs when ${\bf\nabla}\! f$ and the unit vector $\mathbf{u}$ have the same direction. That is, the directional derivative is a maximum when the direction is $\mathbf{u}=\dfrac{4\sqrt{41}}{41}\mathbf{i}-\dfrac{5\sqrt{41}}{ 41}\mathbf{j}$.

(b) The maximum value of the directional derivative at $(1,-2)$ equals the magnitude of the gradient, namely $$\Vert {\bf\nabla}\! f(1,-2)\Vert =\sqrt{16+25}=\sqrt{41}$$