Before you do anything, you have to find [tex]c[/tex] such that [tex]f_{X,Y}(x,y)[/tex] is a proper joint density function. Doing the math, you'll find that [tex]c=2[/tex].
Now, determine the marginal densities:
[tex]f_X(x)=\displaystyle\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dy=\int_{-x}^x2(x^2-y^2)e^{-2x}\,\mathrm dy[/tex]
[tex]\implies f_X(x)=\dfrac83x^3e^{-2x}[/tex]
[tex]f_Y(y)=\displaystyle\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dx=\int_0^\infty2(x^2-y^2)e^{-2x}\,\mathrm dx[/tex]
[tex]\implies f_Y(y)=\dfrac12-y^2[/tex]
a. Then the density of [tex]X[/tex] conditioned on [tex]Y=y[/tex] is
[tex]f_{X\mid Y}(x\mid Y=y)=\dfrac{f_{X,Y}(x,y)}{f_Y(y)}=\dfrac{4(x^2-y^2)e^{-2x}}{1-2y^2}[/tex]
b. The density of [tex]Y[/tex] conditioned on [tex]X=x[/tex] is
[tex]f_{Y\mid X}(y\mid X=x)=\dfrac{f_{X,Y}(x,y)}{f_X(x)}=\dfrac{3(x^2-y^2)}{4x^3}[/tex]
and so the distribution of [tex]Y[/tex] conditioned on [tex]X=x[/tex] is
[tex]F_{Y\mid X}(y\mid X=x)=\displaystyle\int_{-\infty}^uf_{Y\mid X}(y\mid X=x)\,\mathrm du[/tex]
[tex]F_{Y\mid X}(y\mid X=x)=\begin{cases}0&\text{for }y<-x\\\frac{2x^3+3x^2y-y^3}{4x^3}&\text{for }-x\le y\le x\\1&\text{for }y>x\end{cases}[/tex]
