Integer solution $(x,y,z)$ to $xyz=24$ 
Find the total number of integer solutions $(x,y,z)\in\mathbb{Z}$ of the equation $xyz=24$.

I have tried $xyz = 2^3 \cdot 3$
My Process:
Factor $x$, $y$, and $z$ as
$$
\begin{cases}
x = 2^{x_1} \cdot 3^{y_1}\\
y = 2^{x_2} \cdot 3^{y_2}\\
z = 2^{x_3} \cdot 3^{y_3}
\end{cases}
$$
Now $2^{x_1+x_2+x_3} \cdot 3^{y_1+y_2+y_3} = 2^3 \cdot 3$, so
$$
\begin{cases}
x_1+x_2+x_3 = 3\\
y_1+y_2+y_3 = 1
\end{cases}
$$
where $x_1,x_2,x_3,y_1,y_2,y_3\geq 0$. Thus the number of solutions for the first is $\binom{5}{2}=10$ and second is $\binom{3}{2}=3$. This gives a total number of positive solutions of $10 \cdot 3 = 30$
How can I calculate the total number of solutions? Is this process correct?

Yes, this looks right for finding the number of positive integer solutions.
Since you originally asked for all integer solutions, you'll also have to take into account the signs of the factors. There are $\binom{3}{0} + \binom{3}{2} = 4$ ways of adding zero or two minus signs, for a total of 120 integer solutions.
