Find the variance of $Y = X_1 X_2 / X_3$ where $X_i \sim$ Uniform(1,2) for $i = 1,2,3$ The problem is very straightforward and it says:

Let $X_1,X_2,X_3$ independent random variables with a common density $f_{x_i} \sim Uniform(1,2)$. Define $Y = X_1 X_2 / X_3$ Find the numeric value of $var(Y)$.

I propose the following transformation in order to compute the density of $Y_1$ through the margin distribution, and then compute the variance of $Y_1$:
$$ Y_1 = X_1 X_2 / X_3 $$
$$ Y_2 = X_2/X_3 $$
$$ Y_3 = X_3 $$
But I'm not getting any results. Is there an easy path to follow?
Disclaimer: I tried to solve the problem with this technique because I wanted to practice the topic of transformation of random variables.

One easy path is to apply standard definitions.  In particular, when the $X_i$ are independent and identically distributed with a density $f,$ the moment of $Y=X_1X_2/X_3 = y(X_1,X_2,X_3)$ of order $k$ is given by
$$\mu_k(Y) = \iiint y(x_1,x_2,x_3)^kf(x_1)f(x_2)f(x_3)\,\mathrm{d}x_1 \mathrm{d}x_2 \mathrm{d}x_3.$$
For the Uniform$[1,2]$ distribution, $f(x) = 1$ for $1\le x \le 2$ (and otherwise is zero).  The domain of integration thereby reduces to the cube $[1,2]^3,$ giving
$$\mu_k(Y) = \int_1^2\int_1^2\int_1^2 \left(\frac{x_1x_2}{x_3}\right)^k\,\mathrm{d}x_1 \mathrm{d}x_2 \mathrm{d}x_3.$$
The integrand factors, enabling this triple integral to be computed as a product of three elementary univariate integrals,
$$\mu_k(Y) = \int_1^2x_1^k\,\mathrm{d}x_1 \int_1^2x_2^k\,\mathrm{d}x_2\int_1^2 x_3^{-k}\,  \mathrm{d}x_3 = \left(\frac{x_1^{k+1}}{k+1}\bigg|^2_1\right) \left(\frac{x_2^{k+1}}{k+1}\bigg|^2_1\right) \left(\frac{x_3^{1-k}}{1-k}\bigg|^2_1\right)$$
(understanding that any expression evaluating to "$x^0/0$" means $\log(x)$).
Apply this result to the cases $k=1,2$ and plug them into the formula
$$\operatorname{Var}(Y) = \mu_2(Y) - \mu_1(Y)^2.$$
