So law of total variance states that: $$Var(Y) = E[ Var(Y | X)] + Var(E[Y|X])$$ Now it's given that $X,Y$ are positive i.i.d. continuous r.v.s.
As $X,Y$ are independent, it means $E[Y|X] = E[Y]$, which is a constant. Let it be $\mu$.
Then, $$Var(E[Y|X]) = Var(E[Y]) = Var(\mu) = 0$$ Thus, $$Var(Y) = E[ Var(Y | X)] + 0 = E[ Var(Y | X)]$$
Is my reasoning right?
$$\operatorname{Var}\left(Y\right) = \operatorname{E}\left[\operatorname{Var}\left(Y\mid X\right)\right] +\operatorname{Var}\left(\operatorname{E}\left[Y\mid X\right]\right)$$ If $Y$ is independent of $X$, then $$\operatorname{E}\left[\operatorname{Var}\left(Y\mid X\right)\right]=\operatorname{E}\left[\operatorname{Var}\left(Y\right)\right]$$ And $$\operatorname{Var}\left(\operatorname{E}\left[Y\mid X\right]\right)=\operatorname{Var}\left(\operatorname{E}\left[Y\right]\right)=0$$ Therefore $$\operatorname{Var}\left(Y\right) = \operatorname{E}\left[\operatorname{Var}\left(Y\right)\right]$$ Which makes sense because the expected value of a constant is equivalent to the constant. So yes, as others have already mentioned, your reasoning is indeed correct.