I have a region in a 3-D space with a density of $$ \ f_{x,y,z}(X,Y,Z) = \begin{cases} 1 & \text{if $ (x,y,z)\in W$}; \\ 0 & \text{if $(x,y,z)\notin W$};\\ \end{cases} \ $$
Being $W$ the set of points inside the pyramid $(0,0,0)$, $(2,0,0)$,$(0,3,0)$ and $(0,0,1)$
Im asked to find $P\{\frac{1}{3}<Z<\frac{2}{3}\}$
For doing so i find first the plane situated on the front part of the pyramid: $z=1-\frac{y}{3}-\frac{x}{2}$
And i will try to find the marginal density probability of $Z$ to find $P\{\frac{1}{3}<Z<\frac{2}{3}\}$
I know the area of the base of the pyramid by doing $y=0 \rightarrow x=2(1-z) $ and $x=0 \rightarrow y=3(1-z)$. So the area is $\frac{xy}{2}=3(1-z)^2$
How can i manage from here to find $P\{\frac{1}{3}<Z<\frac{2}{3}\}$?
Really, you've done just about all of the work. The only missing part is the easy one. You know the volume of the pyramid is one, so that the probability is the fraction of the volume between $z \in [1/3,2/3]$, or
$$3 \int_{1/3}^{2/3} dz \, (1-z)^2 = \left (\frac23 \right )^3 - \left (\frac13 \right )^3 = \frac{7}{27}$$
Actually, you don't even need integration. You know that the area of the base at $z=2/3$ is $3 (1/3)^2 = 1/3$ and at $z=1/3$ is $3 (2/3)^2 = 4/3$. The volume is then
$$\frac13 \cdot \frac23 \cdot \frac43 - \frac13 \cdot \frac 13 \cdot \frac 13 = \frac{7}{27}$$