I have tried using the formula of the variance, $$V(X) = E\left [ X-E(X) \right ]^{2} = E(X^{2}) - \left [ E(X) \right ]^{2}.$$
So I have $$V\left ( Y\mid X \right ) = E\Bigl (\Bigl \{ E(Y\mid X) - E \bigl (E(Y\mid X) \bigr) \Bigr\} ^{2}\Bigr)$$
and $$E\bigl ( V(Y\mid X) \bigr) = E \Bigl\{ E\Bigl[ \bigl\{ E(Y\mid X)-E(Y\mid X) \bigr\}^{2} \Bigr] \Bigr\}$$.
Not sure how to proceed from here to show how the LHS will hold.
The conditional variance is define as: \begin{align} Var(Y|X) = E[\left(Y-E[Y|X])^2|X\right) \end{align} In one hand, we have \begin{align} Var\left (E\left(Y|X\right)\right) &= E[E\left(Y|X\right)^2] - E[E\left(Y|X\right)]^2\\ &= E[E\left(Y|X\right)^2] - E[Y]^2 \end{align} In other hand, we have \begin{align} E[Var(X|Y)] &= E[E[\left(Y-E[Y|X]\right)^2|X]]\\ &=E[\left(Y-E[Y|X])^2\right)\\ &=E[Y^2] - 2E[YE[Y|X]] +E[E[Y|X]^2]\\ &=E[Y^2] - E[E[Y|X]^2] \end{align} Adding everything we have, \begin{align} Var\left ( Y \right )=Var\left ( E\left ( Y|X \right ) \right )+E\left ( Var\left ( Y|X \right ) \right ) \end{align} Note that $E[YE[Y|X]]=E[E[\left(YE[Y|X]\right)|X]]=E[E[Y|X]E[Y|X]]=E[E[Y|X]^2]$