= Solution
The <gradient> is $\nabla f=(2x+y,x)$ and at $(1,1)$ it is $(3,1)$. For a <unit vector> $\hat{\mathbf n}$, the <directional derivative> is $D_{\hat{\mathbf n}}f=(3,1)\cdot\hat{\mathbf n}$. The <Cauchy-Schwarz inequality> bounds its absolute value by $\sqrt{10}$, with equality in the parallel or antiparallel direction. Thus
$$
\boxed{\hat{\mathbf n}=\pm\frac{(3,1)}{\sqrt{10}},\qquad |D_{\hat{\mathbf n}}f|_{\max}=\sqrt{10}.}
$$
The plus sign gives greatest increase; the minus sign gives greatest decrease, with the same magnitude. Stationary directions at the specified point have zero first derivative, so they are perpendicular to the gradient:
$$
\boxed{\hat{\mathbf t}=\pm\frac{(1,-3)}{\sqrt{10}}.}
$$
These are tangents to the local <level curve>, not a claim that $(1,1)$ is itself a stationary point of the two-variable function.
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