For any two random variables $X,Y$, we can define a random variable $X\vert Y$ which represents the value of $X$ given the value of $Y$. Thus, according to the standard definition, we have $$V(X\vert Y )= E(X^2\vert Y) - E^2 (X\vert Y)$$ The RHS is some function of $Y$ so $V(X\vert Y)$ would be a function of $Y$ which is a random variable and you can calculate it's variance in the usual manner. I don't see a way to provide a better answer, as it really depends on the random variables and their dependence.

For example, let $Y$ be some RV and $X\vert Y$ be either $Y$ or $-Y$ with equal probabilities. Then $E(X\vert Y)=0$ and $E(X^2\vert Y)=Y^2$, hence $Var(X\vert Y)=Y^2$. It follows that $$V(V(X\vert Y)) = V(Y^2)=E((Y^2)^2)-E^2(Y^2)=E(Y^4)-E^2(Y^2)$$ Choose your favorite $Y$, compute the appropriate expectations and you'll have your answer.

Use one of the Definitions for Variance (they are equivalent, so use whichever you find easiest): $$\begin{align}\mathsf{Var}(\mathsf {Var}(X\mid Y))&=\mathsf E(\mathsf{Var}(X\mid Y)^2)-\mathsf E(\mathsf {Var}(X\mid Y))^2\\[1ex] &=\mathsf E\big(\big(\mathsf{Var}(X\mid Y)-\mathsf E(\mathsf{Var}(X\mid Y))\big)^2\big)\end{align}$$

Since you say that you have the joint probability measure function to evaluate, you shall be able to calculate the conditional variance of $X$ under $Y$.

$$\begin{align}\mathsf{Var}(X\mid Y)&=\mathsf E(X^2\mid Y)-\mathsf E(X\mid Y)^2\\[1ex]&=\mathsf E((X-\mathsf E(X\mid Y))^2\mid Y)\end{align}$$

In the continuous case, for $k\in\Bbb N^+$:

$$\mathsf E(X^k\mid Y)= \dfrac{\displaystyle\int_\Bbb R x^kf_{X,Y}(x,Y)~\mathrm d x}{\displaystyle\int_\Bbb R f_{X,Y}(x,Y)~\mathrm d x}$$

And similarly, in the discrete case:

$$\mathsf E(X^k\mid Y)= \dfrac{\sum_{x} x^kf_{X,Y}(x,Y)}{\sum_{x} f_{X,Y}(x,Y)}$$