Answer
${f_x}\left( {1,3} \right) = \frac{7}{8}$ and ${f_y}\left( {1,3} \right) = \frac{3}{4}$.
Work Step by Step
We have $f\left( {x,y} \right) = \sqrt {7x + {y^2}} $. The partial derivatives are
${f_x} = \frac{7}{{2\sqrt {7x + {y^2}} }}$, ${\ \ \ }$ ${f_y} = \frac{y}{{\sqrt {7x + {y^2}} }}$
So,
${f_x}\left( {1,3} \right) = \frac{7}{{2\sqrt {16} }} = \pm \frac{7}{8}$
${f_y}\left( {1,3} \right) = \frac{3}{{\sqrt {16} }} = \pm \frac{3}{4}$
Since $f\left( {x,y} \right)$ is increasing, we choose the positive values. Hence, ${f_x}\left( {1,3} \right) = \frac{7}{8}$ and ${f_y}\left( {1,3} \right) = \frac{3}{4}$.
