$y=\frac{2}{3} x^3+\frac{1}{2} x^2$
$$
\begin{aligned}
\therefore \quad \frac{d y}{d x} & =\frac{2}{3}\left(3 x^2\right)+\frac{1}{2}(2 x) \\
& =2 x^2+x
\end{aligned}
$$
$\because$ Tangents make equal angles with coordinate axis.
Hence, $\frac{d y}{d x}=1$
$$
\begin{aligned}
& \Rightarrow 2 x^2+x=1 \\
& \Rightarrow \quad 2 x^2+2 x-x-1=0 \\
& 2 x(x+1)-1(x+1)=0 \\
& \Rightarrow \quad(x+1)(2 x-1)=0 \Rightarrow x=\frac{1}{2},-1 \\
&
\end{aligned}
$$
At
$$
\begin{aligned}
x & =\frac{1}{2}, y=\frac{2}{3}\left(\frac{1}{2}\right)^3+\frac{1}{2}\left(\frac{1}{2}\right)^2 \\
& =\frac{1}{12}+\frac{1}{8}=\frac{5}{24} \\
y & =\frac{5}{24}
\end{aligned}
$$

At $x=-1, y=\frac{2}{3}(-1)^3+\frac{1}{2}(-1)^2=\frac{-2}{3}+\frac{1}{2}$
$$
\begin{gathered}
y=\frac{-1}{6} \\
\therefore\left(\frac{1}{2}, \frac{5}{24}\right) \text { and }\left(-1, \frac{-1}{6}\right)
\end{gathered}
$$