I'll assume you're supposed to compute the line integral of [tex]\nabla f[/tex] over the given path [tex]C[/tex]. By the fundamental theorem of calculus,
[tex]\displaystyle\int_C\nabla f(x,y)\cdot\mathrm d\vec r=f(4,48)-f(1,3)[/tex]
so evaluating the integral is as simple as evaluting [tex]f[/tex] at the endpoints of [tex]C[/tex]. But first we need to determine [tex]f[/tex] given its gradient.
We have
[tex]\dfrac{\partial f}{\partial x}=3y\sin(xy)\implies f(x,y)=-3\cos(xy)+g(y)[/tex]
Differentiating with respect to [tex]y[/tex] gives
[tex]\dfrac{\partial f}{\partial y}=3x\sin(xy)=3x\sin(xy)+\dfrac{\mathrm dg}{\mathrm dy}\implies\dfrac{\mathrm dg}{\mathrm dy}=0\implies g(y)=C[/tex]
and we end up with
[tex]f(x,y)=-3\cos(xy)+C[/tex]
for some constant [tex]C[/tex]. Then the value of the line integral is [tex]-3\cos192+3\cos3[/tex].
