So, I have the following problem:
Find $\frac{\partial z}{\partial x}$ if $x^2-3yz^2+xyz-2=0$.
It seems to me that this problem is meant to be solved using implicit differentiation, except I'm a bit confused that the question asks for the partial derivative instead of just $\frac{dz}{dx}$; does this distinction change the answer?
Just performing implicit differentiation, I get
$\frac{\partial z}{\partial x} = \frac{-2x+3\frac{dy}{dx}z^2-yz-x\frac{dy}{dx}z}{6yz+xy}$
If possible, could someone tell me whether this is correct?
Yes.
Partial derivatives of an expression are taken with respect to one of a set of independent variables. So what are the dependent and independent variables here?
If we consider $y$ and $z$ both dependent on $x$, then you get your result. But this would be the monovariate derivative: $\frac{\mathrm d z}{\mathrm d x}$.
So, since you were asked for a partial derivative, this suggests they intended for you to consider $z$ implicitly dependent on $x$ and $y$.
Thus you should have taken the partial difference of everything with respect to $x$ independent of $y$. This is sometimes clarified by the addition of a subscript.
$$x^2−3yz^2+xyz−2=0\\\dfrac{\partial x^2}{\partial x}_{\vert y}-3\dfrac{\partial (yz^2)}{\partial x}_{\vert y}+\dfrac{\partial (xyz)}{\partial x}_{\vert y}=0\\2x-6yz\dfrac{\partial z}{\partial x}_{\vert y}+\left(xy\dfrac{\partial z}{\partial x}_{\vert y}+yz\right)=0\\\dfrac{\partial z}{\partial x}_{\vert y}=\dfrac{-(2x+yz)}{-6yz+xy}$$
This is identical to your result, except that the derivatives of $y$ vanish. $\left[\frac{\partial y}{\partial x}_{\vert y}=0\right]$