Proof: There is no function $f \in C^2(\mathbb{R^3})$ with gradient $\nabla f(x,y,z) = (yz, xz, xy^2)$

Question

How can one show that there is no function, which is a continuously partially derivable function $f \in C^2(\mathbb{R^3})$ with this gradient

$$\nabla f(x,y,z) = (yz, xz, xy^2)$$

I thought about using the Hessian matrix since one has to calculate all second partial derivatives of $f$ there.

Since only the gradient is given, can I calculate the antiderivatives first:

$yz = xyz$

$xz = xyz$

$xy^2 = xy^2z$

Now I want to calculate the antiderivatives of the antiderivatives:

$xyz = \frac{yzx^2}{2}$

$xyz = \frac{yzx^2}{2}$

$xy^2z = \dfrac{y^2zx^2}{2}$

I didn't calculate the antiderivatives of the partial derivatives and I don't even know if that way is correct...

Answer 1

If $$\nabla f=(f_1(x,y,z),f_2(x,y,z),f_3(x,y,z))$$and $f_i$s are twice differentiable, we must have $${\partial ^2 f_1\over \partial y\partial z}={\partial ^2 f_2\over \partial x\partial z}={\partial ^2 f_3\over \partial x\partial y}$$for all $x,y,z$ but this doesn't hold here since $$1=1\ne 2y\qquad \forall x,y,z\in \Bbb R$$therefore such a function doesn't exist.

Answer 2

In other words, you are asking to show that your vector field is not conservative. It is enough to show that the curl of your vector field is different to zero.