Let $z=x+i y$
Now, we have $\quad|z|+|z-1|=3$
$$
\begin{array}{lc}
\Rightarrow & |x+| y|+| x+|y-1|=3 \\
\Rightarrow & \sqrt{x^2+y^2}+\sqrt{(x-1)^2+y^2}=3 \\
\Rightarrow & \sqrt{(x-1)^2+y^2}=3-\sqrt{x^2+y^2} \\
\Rightarrow & (x-1)^2+y^2=9+x^2+y^2-6 \sqrt{x^2+y^2} \\
\Rightarrow & x^2-2 x+1+y^2=9+x^2+y^2-6 \sqrt{x^2+y^2}
\end{array}
$$

$$
\begin{array}{lrl}
\Rightarrow & & -2 x+1=9-6 \sqrt{x^2+y^2} \\
\Rightarrow & & 6 \sqrt{x^2+y^2}=2 x+8 \\
\Rightarrow & 3 \sqrt{x^2+y^2}=x+4 \\
\Rightarrow & 9\left(x^2+y^2\right)=(x+4)^2 \\
\therefore & 9 x^2+9 y^2=x^2+8 x+16 \\
\Rightarrow & 8 x^2+9 y^2-8 x=16 \\
\Rightarrow & 8\left(x^2-x\right)+9 y^2=16 \\
\Rightarrow & 8\left[\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\right]+9 y^2=16
\end{array}
$$

$$
\begin{array}{lrl}
\Rightarrow & 8\left(x-\frac{1}{2}\right)^2-2+9 y^2=16 \\
\Rightarrow & 8\left(x-\frac{1}{2}\right)^2+9 y^2=18 \\
\therefore & & \frac{\left(x-\frac{1}{2}\right)^2}{\frac{9}{4}}+\frac{y^2}{2}=1
\end{array}
$$
Which is an equation of ellipse.