This is a problem that can be found in the text High-Dimensional Statistics: A Non-Asymptotic Viewpoint by Martin Wainwright. An important hint is the first part of the question asking you to prove that $\phi'(z) + z\phi(z) = 0$, which comes directly from computing $\phi'(z) = \frac{-z}{\sqrt{2\pi}} e^{-z^2/2}$.
Using the above, we may note first that $1 - \Phi(z) = \mathbb{P}[Z\geq z] = \int^\infty_z \phi(t) dt$, and substituting $\phi(z) = \frac{-\phi'(z)}{z}$ and using integration by parts ($u = \frac{1}{t}$, $dv=\phi'(t)$) we get
$$
\int^\infty_z \phi(t) dt = \int^\infty_z \frac{-\phi'(t)}{t} dt = \left[ \frac{-\phi(t)}{t}\right]^\infty_z - \int^\infty_z \frac{\phi(t)}{t^2} dt
$$
Since $\lim_{t\to \infty} \frac{-\phi(t)}{t} = 0$, we get the top as $\frac{\phi(z)}{z} - \int_z^\infty \frac{\phi(t)}{t^2} dt$, where we may use the same substitution and apply integration by parts again:
$$
\frac{\phi(z)}{z} - \int_z^\infty \frac{-\phi'(t)}{t^3} dt = \frac{\phi(z)}{z} + \left[ \frac{\phi(t)}{t^3}\right]^\infty_z - \int^\infty_z \frac{-3\phi(t)}{t^4}dt
$$
$$
= \frac{\phi(z)}{z} - \frac{\phi(z)}{z^3} + \int^\infty_z \frac{3\phi(t)}{t^4}dt
$$
Thus since $\int^\infty_z \frac{3\phi(t)}{t^4}dt>0$ we get $\phi(z)\left(\frac{1}{z} - \frac{1}{z^3}\right) < \mathbb{P}[Z\geq z]$.
Applying the trick again to $\int^\infty_z \frac{3\phi(t)}{t^4}dt$ yields
$$
\int^\infty_z \frac{-3\phi'(t)}{t^5}dt = \left[ \frac{-3\phi(t)}{t^5}\right]^\infty_z - \int^\infty_z \frac{15\phi(t)}{t^6}dt
$$
$$
= \frac{3\phi(z)}{z^5} - \int^\infty_z \frac{15\phi(t)}{t^6}dt
$$
and since $- \int^\infty_z \frac{15\phi(t)}{t^6}dt<0$, we get $\mathbb{P}[Z\geq z] < \phi(z)\left(\frac{1}{z} - \frac{1}{z^3} + \frac{3}{z^5}\right) $, which proves the claim.
I believe its straightforward to derive a similar bound for $\Phi(X)$ for $x<0$.
