Find the conditional probability density function of $X$ given $Y=y$ Let $X$ and $Y$ be two random variables with joint density function:  $f_{XY}(x,y) = \begin{cases} {5x^2y}&-1\leq x\leq1, 0<y\leq|x| \\{0}&\text{otherwise} \end{cases}$
Find $f_{X|Y}(x|y)$ the conditional probability density function of $X$ given $Y =
y$. Sketch the graph of $f_{X|Y}(x|.5)$
The graph part isn't the confusing issue as I feel once I get to it, I will be able to solve it. My issue is finding the proper $f_{X|Y}(x|y)$. I thought I would do $f_y(y)=\int_{-1}^{1}5x^2ydx$ and I get $10y\over3$ then I plug that into the formula to get $5x^2y\over{10y\over3}$ $= {3x^2\over2}$ However, the answer is ${3x^2\over{2(1-y^3)}}$ What am I doing wrong? I feel like it has to do with my limits of integration but I don't see how it isn't $-1$ to $1$.

First, sketch the support of $(X,Y)$, which is $$(-1 \le X \le 1) \cap (0 < Y \le |X|).$$  This describes a "bow-tie" shaped region in the Cartesian coordinate plane, consisting of two right triangles with vertices $$\{(0,0), (1,0), (1,1)\}, \quad \{(0,0), (-1,0), (-1,1)\}.$$  The symmetry of this region is clear, and the line of symmetry is the $y$-axis.

The plot below visualizes the joint density in three-dimensional space:

For a fixed $Y = y$, the permissible values of $X$ comprise the union of two intervals; that is to say, $$(X \mid Y = y) \in [-1, -y] \cup [y, 1].$$  So, the interval of integration for the marginal density of $Y$ is not $[-1,1]$ but the interval above:  $$f_Y(y) = \int_{x=-1}^{-y} 5x^2 y \, dx + \int_{x=y}^1 5x^2 y \, dx.$$  

The above animation plots $X$ for specific values of $Y = y$ for $y \in (0,1]$.  The conditional density $f_{X\mid Y}(x \mid y)$ is proportional to this plot in such a way that the area under the curve is made equal to $1$.  In other words, the function $f_{X \mid Y}$ is simply a scaled version of this animation such that the area under the curve remains $1$ regardless of the choice of $y$:

Compare this with the previous animation.
Another way to reason about this is to observe that the support requires that $|X| \ge Y$:  hence, on the interval $X \in [-y,y]$, $f_{X,Y}(x,y) = 0$, thus writing $$f_Y(y) = \int_{x=-1}^1 f_{X,Y}(x,y) \, dx = \int_{x=-1}^1 5x^2 y \, dx$$ fails to reflect that the joint density is not $5x^2 y$ when $-y \le x \le y$.  We can correct for this by the suitable use of indicator functions; for example, we could have instead written the joint density as $$f_{X,Y}(x,y) = 5x^2 y \;\mathbb 1(|x| \ge y), \quad (x,y) \in [-1,1] \times [0,1].$$  This reminds us that integration in the rectangular region $[-1,1] \times [0,1]$ comes with an extra condition for the integrand to be positive:  $$f_Y(y) = \int_{x=-1}^1 5x^2 y \; \mathbb 1(|x| \ge y) \, dx = \int_{x=-1}^{-y} 5x^2 y \, dx + \int_{x=y}^1 5x^2 y \, dx.$$  We could go overboard and write the whole density as $$f_{X,Y}(x,y) = 5x^2 y \; \mathbb 1 (0 < y \le |x| \le 1), \quad (x,y) \in \mathbb R^2$$ but this isn't really necessary as we already intuitively grasp that the convex boundary of the support is rectangular, leading to a natural choice of the lower and upper limits of integration except when dealing with the nonconvex part.
