Given, $y=x^3-3 x^2+5$
On differentiating both sides w.r.t. ' $x$ ', we get
$$
\frac{d y}{d x}=3 x^2-6 x
$$
For local maxima or local minima, put $\frac{d y}{d x}=0$
$$
\begin{aligned}
& \Rightarrow & 3 x^2-6 x & =0 \\
& \Rightarrow & 3 x(x-2) & =0 \\
& \Rightarrow & x & =0 \text { or } x=2
\end{aligned}
$$
Now, differentiating Eq. (ii) w.r.t. ' $x$ ' we get
$$
\begin{aligned}
& \frac{a^2 y}{d x^2}=6 x-6 \\
& \Rightarrow \quad\left(\frac{a^2 y}{d x^2}\right)_{x=0}=-6 < 0 \\
& \therefore x=0 \text { is a point of local maxima. } \\
& \text { and }\left(\frac{a^2 y}{d x^2}\right)_{x=2}=6 \times 2-6=12-6=6>0 \\
& \therefore x=2 \text { is a point of local minima. }
\end{aligned}
$$