Answer
$\color{blue}{f_X(x) = 3(1-x)^2,\ 0\le x\le 1}$
Work Step by Step
Using Theorem 3.7.2 (p. 167):
$\begin{align*}
f_X(x) &= \int_{\Omega_Y} f_{X,Y}(x,y)\ dy,\ 0\le x,y\le 1,\ 0\le x + y\le 1 \\
&= \int_{\Omega _Y} 6(1-x-y)\ dy,\underbrace{\ 0\le y\le 1-x}_{\Omega_Y},\ 0\le x\le 1 \\
&= \int_0^{1-x} 6(1-x-y)\ dy,\ 0\le x\le 1 \\
&= 6\left((1-x)y - \frac{y^2}{2}\ \right\vert_{y=0}^{y=1-x} \\
&= 6\biggl( \left((1-x)(1-x) - \frac{(1-x)^2}{2}\right) - \left( (1-x)(0) - \frac{0^2}{2}\right) \biggr) \\
&= 6\left( (1-x)^2 - \frac{(1-x)^2}{2} -0 -0 \right) \\
&= 6\frac{(1-x)^2}{2} \\
\color{blue}{f_X(x)}\ &\color{blue}{= 3(1-x)^2,\ 0\le x\le 1.}
\end{align*}$
