Poisson Process: Density of Quotient The question is:

Let $\{N(t): t \geq 0\}$ be a Poisson process of rate $1$ and let $T_1 < T_2 < \cdots$ denote the times of the points. Then what is the pdf of $Y = \frac{T_1}{T_3}?$

What I know:
$P(T_1 \leq t) = P(N(t) \geq 1) = 1 - P(N(t) = 0) = 1 - e^{-t} \implies$ pdf of $T_1$ is $e^{-t}$
$P(T_3 \leq t) = 1 - P(N(t) = 0)-P(N(t) = 1)-P(N(t) = 2) = 1 - e^{-t} - te^{-t} - t^2e^{-t} \implies$ pdf of $T_3$ is $t^2e^{-t}-te^{-t}$
But I don't recall that there is a formula where if you have the density of two random variables, then the density of their quotient is simply the quotient of their densities so what is the next step here or the correct way to do this problem?
 A: Since $\left\{N(t);t\ge 0\right\}$ is a Poisson process of rate $1$, by definition $T_k-T_{k-1}$ are i.i.d$\sim\exp(1)\ \forall k\ge 1$ with $T_0=0$. Hence, the question boils down to finding the pdf of $X=\frac{X_1}{X_1+X_2+X_3}$ where $X_1,X_2,X_3$are i.i.d.$\sim\exp(1)$. 
Now, note that $X_2+X_3\sim \mbox{Erlang}(2,1)$ where a RV $X\sim \mbox{Erlang}(k,\lambda)$ means that the pdf of $X$ is given by $$f(x;k,\lambda)=\frac{\lambda^kx^{k-1}e^{-\lambda x}}{\Gamma(k)}$$ for $x,\lambda\ge 0$. Also note that $$X_1\perp X_2+X_3$$ Then, clearly  \begin{equation}\begin{split} 
\mathbb{P}(X\le x)=1
\end{split}\end{equation} for $x\ge 1$ and for $x\in [0,1)$\begin{equation}\begin{split} 
\mathbb{P}(X> x)=&\int_{0}^{\infty}\mathbb{P}\left(X_1>\frac{xu}{1-x}\Big|X_2+X_3=u\right)f_{X_2+X_3}(u)du\\
\ =&\int_{0}^{\infty}\mathbb{P}\left(X_1>\frac{xu}{1-x}\right)ue^{-u}du\\
\ =&\int_{0}^{\infty}e^{-\frac{xu}{1-x}}ue^{-u}du=\int_{0}^{\infty}ue^{-\frac{u}{1-x}}du=(1-x)^2\\
\end{split}\end{equation}
$$\therefore \mathbb{P}(X\le x)=\left\{\begin{array}{rl}
0 & \mbox{if}\ x\le 0\\
1 & \mbox{if} \ x\ge 1\\
1-(1-x)^2 & \mbox{if}\ x\in (0,1)
\end{array}
\right.
$$
Hence $$f_X(x)=\frac{d}{dx}\mathbb{P}(X\le x)=\left\{\begin{array}{rl}
2(1-x) & \mbox{if}\ x\in (0,1)\\
0 & \mbox{else}
\end{array}
\right.
$$
