\(f(x,y){\rm{ }} = {\rm{ }}x{e^{xy}}\)
\(\overline V f\left( {x,y} \right) = {f_x}\left( {x,y} \right)i + {f_y}\left( {x,y} \right)j\)
Applying the formula to the given function and using summation derivation rule, we get
\(\begin{aligned}\overline V f\left( {x,y} \right) &= {f_x}\left( {x,y} \right)i + {f_y}\left( {x,y} \right)j \\ &= \frac{\partial }{{\partial x}}\left( {x{e^{xy}}} \right)i + \frac{\partial }{{\partial y}}\left( {x{e^{xy}}} \right)j \\ &= \frac{\partial }{{\partial x}}\left( {\mathop{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{x} }\limits_u \mathop {{{\mathop{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{e} }\limits_v}^{xy}}}\limits_{} } \right)i + x\frac{\partial }{{\partial y}}\left( {{e^{xy}}} \right)j\\&= \left( {1 \cdot {e^{xy}} + x \cdot y{e^{xy}}} \right)i + x\frac{\partial }{{\partial y}}\left( {{e^{xy}}} \right)j \\&= {e^{xy}}\left( {1 + xy} \right)i + \left( {{x^2}{e^{xy}}}\right)j \\\end{aligned} \)
