Proving that $\vec F$ is conservative field

Question

I need to prove that $\vec F$ is coservative field:

$$\vec F=\underbrace{\bigg(yz+\frac{1}{yz} \bigg)}_{Q} \hat i+\underbrace{\bigg(xz-\frac{x}{y^2z} \bigg)}_{P}\hat j+\underbrace{\bigg(xy-\frac{x}{yz^2} \bigg)}_R\hat k$$

My attempt:

$\vec F$ is conservative field iff $\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}=\frac{\partial R}{\partial z}$

but $\frac{\partial Q}{\partial x} \neq \frac{\partial P}{\partial y}$

Answer 1

As James said, you want to find a potential function. A field $\vec{F}$ is conservative if it can be written as $\vec{\nabla}V$. In that case you have $$ \vec{F} = P \hat{i} + Q \hat{j} + R \hat{k} = \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z}\hat{k}$$

The way you wrote the conservative field condition is wrong for the order of terms. If defined $P$, $Q$, $R$ as here, the condition is $\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}$, and is equivalent to requiring that $\tfrac{\partial^2 V}{\partial x \partial y} = \tfrac{\partial^2 V}{\partial y \partial x}$.

Answer 2

$V(x,y,z)=xyz+\frac{x}{yz}$(+ a constant) is the potential $\Rightarrow$ the field is conservative.