Answer
$\lt \frac{1}{\sqrt {2s+3t}},\frac{3}{2\sqrt {2s+3t}}\gt$
Work Step by Step
$f(s,t)=\sqrt {2s+3t}$
$f_s(s,t)=\frac{1}{\sqrt {2s+3t}}$
$f_t(s,t)=\frac{3}{2\sqrt {2s+3t}}$
Gradient vector field of $f$ is: $\lt \frac{1}{\sqrt {2s+3t}},\frac{3}{2\sqrt {2s+3t}}\gt$
