To find the marginal probability density function (PDF) \( f_X(x) \) of the random variable \( X \), we need to integrate the joint PDF \( f_{X,Y}(x,y) \) over the appropriate range of \( Y \).
Given that the joint PDF \( f_{X,Y}(x,y) = 3 \) on the region defined by \( 0 < x \leq 1 \), \( 0 \leq y \leq 1 \), and \( y \leq x^2 \), we first need to determine the limits for \( y \) when we are conditioning on \( X = x \).
For a fixed \( x \), \( y \) can vary from \( 0 \) to \( x^2 \) (according to the constraints of the region). Therefore, the marginal PDF of \( X \) is calculated as follows:
\[ f_X(x) = \int_{y=0}^{y=x^2} f_{X,Y}(x,y) , dy \]
Substituting \( f_{X,Y}(x,y) \):
\[ f_X(x) = \int_{0}^{x^2} 3 , dy \]
Calculating this integral:
\[ f_X(x) = 3 \int_{0}^{x^2} dy = 3 [y]_{0}^{x^2} = 3(x^2 - 0) = 3x^2 \]
Thus, the marginal PDF of \( X \) is:
\[ f_X(x) = 3x^2 \quad \text{for } 0 < x \leq 1 \]
Now, we can find \( f_X(0.5) \):
\[ f_X(0.5) = 3(0.5^2) = 3(0.25) = 0.75 \]
Therefore, the value of \( f_X(0.5) \) is
\[ \boxed{0.75} \]