I know I can check whether a given (continous-differentiable) vector field (with simple connected domain) is conservative by checking if its curl is zero. In the 2d case for example
$$ \begin{bmatrix} z_1 \\ z_2 \end{bmatrix} = \begin{bmatrix} F_1(x, y) \\ F_2(x, y) \end{bmatrix} $$
so if the curl
$$ \frac{\partial }{\partial x} F_2(x,y) - \frac{\partial }{\partial y} F_1(x,y) = 0 $$
then the vector field is conservative.
But what can be done if the vector field is not available as a set of explicit functions but instead only implicitly (and it is not possible to isolate the expressions)? I.e. like this:
$$ \begin{bmatrix} G_1(x, y, z_1) \\ G_2(x, y, z_2) \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \tag{1} $$
Or even like this:
$$ \begin{bmatrix} H_1(x, y, z_1, z_2) \\ H_2(x, y, z_1, z_2) \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \tag{2} $$
Question: Can I still somehow check if the implicit vector field $(1)$ or even $(2)$ is conservative?
Example: I will give an explicit example:
$$ \begin{align} z_1 &= G_1(x, y, z_1) = x^2 + y^2 + z_1^3 = 0 \\ z_2 &= G_2(x, y, z_2) = (x + y + z_2)^3 = 0 \end{align} $$
Implicit differentiation:
$$ \begin{align} 0 &= 2 y + 3 z_1^2 \frac{\partial z_1}{ \partial y} \\ 0 &= 3(x + y + z_2)^2 \Big(1 + \frac{\partial z_2}{\partial x} \Big) \end{align} $$
Solve for partial derivatives:
$$ \begin{align} \frac{\partial z_2}{\partial x} &= -1 \\ \frac{\partial z_1}{\partial y} &= -\frac{2 y}{3 z_1^2} \end{align} $$
Then the curl should be
$$ \frac{\partial z_2}{\partial x} - \frac{\partial z_1}{\partial y} = \frac{2 y}{3 z_1^2} - 1 \neq 0 $$
So in this case, the vector field should be not a conservative vector field. Is this conclusion and the derivation correct like this?
Yes, with the conditions necessary for the implicit function theorem. This is the generality I think you are looking for and mimics your computation with the explicit example.
Your goal is to compute $$ \frac{\partial z_2}{\partial x}- \frac{\partial z_1}{\partial y}$$ And as long as $\frac{\partial G_1}{\partial z_1}\ne 0$ and $\frac{\partial G_2}{\partial z_2}\ne 0$, $$ \frac{\partial z_2}{\partial x}- \frac{\partial z_1}{\partial y}\\ =\frac{\frac{\partial G_1}{\partial y}}{\frac{\partial G_1}{\partial z_1}}- \frac{\frac{\partial G_2}{\partial x}}{\frac{\partial G_2}{\partial z_2}} $$ Which are in general more approachable to compute.