Using theorem 4 in the source you posted,
$$f_{Y_1Y_2Y_3}(y_1,y_2,y_3) = f_{X_1X_2X_3}(x_1,x_2,x_3)\lvert J \rvert$$
where
$$
J =
\begin{vmatrix}
\frac{\partial x_1}{\partial y_1} & \frac{\partial x_1}{\partial y_2} & \frac{\partial x_1}{\partial y_3}\\
\frac{\partial x_2}{\partial y_1} & \frac{\partial x_2}{\partial y_2} & \frac{\partial x_2}{\partial y_3}\\
\frac{\partial x_3}{\partial y_1} & \frac{\partial x_3}{\partial y_2} & \frac{\partial x_3}{\partial y_3}
\end{vmatrix}
$$
Asumming that $X_1$, $X_2$ and $X_3$ are independent,
$$f_{X_1X_2X_3}(x_1,x_2,x_3) = e^{-(x_1+x_2+x_3)} = e^{-y_3} \qquad \text{for }y_3>0$$
and as you said, $\lvert J \rvert = y_2y_3^2$. Therefore,
$$f_{Y_1Y_2Y_3}(y_1,y_2,y_3) = y_2y_3^2e^{-y_3} \qquad \text{for }y_3>0, 0\lt y_1,y_2 \lt 1$$
To better interpret this result, we can write
$$f_{Y_1Y_2Y_3}(y_1,y_2,y_3) = f_{Y_2Y_3\mid Y_1}(y_2,y_3\mid y_1)f_{Y_1}(y_1) $$
And from this we can see that
$$f_{Y_2Y_3\mid Y_1}(y_2,y_3\mid y_1) = f_{Y_2Y_3}(y_2,y_3) = y_2y_3^2e^{-y_3} \qquad \text{for }y_3>0, 0\lt y_2 \lt 1$$
and
$$f_{Y_1}(y_1) = 1\qquad \text{for }0\lt y_1 \lt 1$$
The marginal of $Y_3$ can be calculated by integrating $f_{Y_2Y_3}(y_2,y_3)$ over $y_2$,
$$f_{Y_3}(y_3) = \int_0^1y_2y_3^2e^{-y_3}dy_2 = \frac{1}{2}y_3^2e^{-y_3}\qquad \text{for }y_3>0$$
In the same way,
$$f_{Y_3}(y_3) = \int_0^\infty y_2y_3^2e^{-y_3}dy_3 = 2y_2\qquad \text{for }0\lt y_2 \lt 1$$
Informal note about the limits of integration:
For $Y_1$: By definition,
$$y_1 = \frac{x_1}{x_1 + x_2}.$$
Since $x_1,x_2 \gt 0$, $y_1 \gt 0$. In addition, the denominator is always greater than the numerator, then $y_1 \lt 1$, which give us $0 \lt y_1 \lt 1$.
For $Y_2$: Same argument than for $Y_1$
For $Y_3$: $y_3 \gt 0$ since it is the sum of positive numbers.
