Answer
$\color{blue}{h(y) = \lambda,\ y\ge 0}$
Work Step by Step
From Question 3.4.8 (p. 136), for a random variable $Y$ that has an exponential pdf (with parameter $\lambda$), the pdf and cdf are, respectively:
$\begin{align*}
f_Y(y) &= \lambda e^{-\lambda y},\ y\ge 0 \\
F_Y(y) &= \begin{cases} 0, & y \lt 0 \\ 1 - e^{\lambda y}, & y \ge 0 \end{cases}
\end{align*}$
Thus, the hazard rate (defined for $y\ge 0$ as both pdf and cdf are defined for these values of $y$) is
$\begin{align*}
h(y) &= \dfrac{f_Y(y)}{1-F_Y(y)} \\
&= \dfrac{\lambda e^{-\lambda y}}{1-(1-e^{-\lambda y})},\ y\ge 0\\
&= \dfrac{\lambda e^{-\lambda y}}{e^{-\lambda y}} \\
\color{blue}{h(y)} &\color{blue}{= \lambda,\ y\ge 0}
\end{align*}$
That is, if $Y$ has an exponential distribution with parameter $\lambda$, then hazard rate is constant and is equal to $\lambda$.
