probability - Set $Z=X+Y$ and calculate the density function $f_Z$ for Z. Given solution doesn't match mine. (Continuous R.V)

I'm having difficulties with task 2, since given solution doesn't equal mine:

$f_{XY}(x,y) = \begin{cases} \left (6·e^{-3x}·e^{-2y} \right) &
\text{if } 00 & \text{otherwise } %
\end{cases}$

Task 1: Show that X and Y are independent

Task 2: Set $Z=X+Y$ and calculate the density function $f_Z$ for Z

I'll post the entire task, since others might seek help for a similar problem.

Task 1:

To show this we use that "Two continuous random variables X and Y are independent if:

$f_{XY}(x,y)=f_X(x)·f_Y(y),\space for\space all \space x,y$

We find the marginal PDF's

$f_X(x)=\int_0^\infty6·e^{-3x}·e^{-2y}$dy

$f_Y(y)=\int_0^\infty6·e^{-3x}·e^{-2y}$dx

We multiply the two:

$f_Y(y)·f_X(x)=6·e^{-3x}·e^{-2y}$

Which is the same as the joint PDF. We conclude they are independent.

Task 2:

We solve it by finding the CDF and differentiate this to find the PDF:

$P(Z\leq z)$

$P(x+y\leq z)$

$P(y\leq z-x)$

We solve the double integral:

$\int_0^\infty \int_0^{(z-x)}(6·e^{-3x}·e^{-2y}dydx=(1-3·e^{-2·z})$

We now have the CDF:

$F_{Z}(z) = \begin{cases} \left (0 \right) &
\text{if } 0>z \\(1-3·e^{-2·z}) & \text{}01 & \text{}z> \infty %
\end{cases}$

To find the PDF we differentiate the CDF:

$\frac{d}{dz}(1-3·e^{-2·z})=6·e^{-2z} $

Giving us a PDF of:

$f_{Z}(z) = \begin{cases} \left (6·e^{-2z} \right) &
\text{if } 00 & \text{otherwise } %

\end{cases}$

However the solution provided is:

$f_Z(z)=6·e^{-2z}·(1-e^{-z})$

What am I doing wrong?

Besides this im unsure of:

The intervals in the CDF

Answer

$Y = z-X = -X + z$ is a linear equation with $y$-intercept $z$ and slope $-1$. The region where $Y \leq -X + z$ is in green below. Note where the line intersects with the $X$ and $Y$ axes.

Thus, the CDF is, according to the graph above,
$$\int_{0}^{z} \int_{0}^{-x+z}6e^{-3x}e^{-2y}\text{ d}y\text{ d}x = 2e^{-3z}-3e^{-2z}+1$$
from which we obtain
$$f_{Z}(z) = -6e^{-3z}+6e^{-2z}=6e^{-2z}-6e^{-3z}=6e^{-2z}(1-e^{-z})$$

for $z > 0$ as desired.