Determine whether or not $\mathbf{F}$ is a conservative vector field. If it is, find a function $f$ such that $\mathbf{F}=\nabla f$. (If the vector field is not conservative, enter DNE.)
\[
\mathbf{F}(x, y)=\left(y^{3} \cos (x)+\cos (y)\right) \mathbf{i}+\left(3 y^{2} \sin (x)-x \sin (y)\right) \mathbf{j}
\]
\[
f(x, y)=
\]

Step 1 ：Given the vector field \(\mathbf{F}(x, y)=(y^{3} \cos (x)+\cos (y)) \mathbf{i}+(3 y^{2} \sin (x)-x \sin (y)) \mathbf{j}\), we need to determine whether it is conservative and find a function \(f\) such that \(\mathbf{F}=\nabla f\) if it is conservative.

Step 2 ：A vector field \(\mathbf{F}\) is conservative if it can be written as the gradient of a scalar function \(f\). This means that the components of \(\mathbf{F}\) are the partial derivatives of \(f\) with respect to the variables.

Step 3 ：We can check if \(\mathbf{F}\) is conservative by checking if the mixed partial derivatives of \(F_1\) and \(F_2\) are equal, i.e., if \(\frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x}\).

Step 4 ：Calculating the mixed partial derivatives, we find that \(\frac{\partial F_1}{\partial y} = 3y^{2}\cos(x) - \sin(y)\) and \(\frac{\partial F_2}{\partial x} = 3y^{2}\cos(x) - \sin(y)\).

Step 5 ：Since the mixed partial derivatives of \(F_1\) and \(F_2\) are equal, the vector field \(\mathbf{F}\) is conservative.

Step 6 ：Now, we can find the function \(f\) by integrating \(F_1\) with respect to \(x\) and \(F_2\) with respect to \(y\). We need to make sure that the constant of integration is the same for both integrals.

Step 7 ：Integrating, we find that \(f(x, y) = x \cos(y) + y^{3} \sin(x)\).

Step 8 ：Final Answer: The function \(f\) such that \(\mathbf{F}=\nabla f\) is \(\boxed{x \cos(y) + y^{3} \sin(x)}\). Therefore, the vector field \(\mathbf{F}\) is conservative.