Given the following joint density function:
$$ \displaystyle f(x,y) = \begin{cases} 0.5 & \text{if $0 \leq x < \frac{1}{3}, \> \frac{1}{3} \leq y < 1, $} \\ 0.75 & \text{if $\frac{1}{3} \leq x < \frac{2}{3}, \> \frac{1}{3} \leq y < 1, $} \\ 0.25 & \text{if $\frac{2}{3} \leq x < 1, \> \frac{1}{3} \leq y < 1, $} \\ 2 & \text{if $0 \leq x \leq \frac{1}{3}, \> 0 \leq y < \frac{1}{3}, $} \\ 3 & \text{if $\frac{1}{3} \leq x < \frac{2}{3}, \> 0 \leq y < \frac{1}{3}, $} \\ 1 & \text{if $\frac{2}{3} \leq x < 1, \> 0 \leq y < \frac{1}{3}, $} \\ \end{cases} $$
The marginal probability is
$$ f_X(x)=\int_{y} f(x,y)dy \\f_Y(y)=\int_{x} f(x,y)dx $$
But if $f(x,y)$ has multiple values for a given $x,y$, how do you evaluate the integral?
To get you "on the road" I will provide one of the cases.
Let it be that $\frac13\leq x<\frac23$.
Then:$$f_{X}\left(x\right)=\int f_{X,Y}\left(x,y\right)\;dy=\int_{0}^{\frac{1}{3}}f_{X,Y}\left(x,y\right)\;dy+\int_{\frac{1}{3}}^{1}f_{X,Y}\left(x,y\right)\;dy=\int_{0}^{\frac{1}{3}}3\;dy+\int_{\frac{1}{3}}^{1}0.75\;dy=$$$$\left[3y\right]_{0}^{\frac{1}{3}}+\left[0.75y\right]_{\frac{1}{3}}^{1}=3\left(\frac{1}{3}-0\right)+0.75\left(1-\frac{1}{3}\right)=1.5$$