At what points of curve $\mathrm{y}=\frac{2}{3}{\mathrm{x}}^3+\frac{1}{2}{\mathrm{x}}^2$ the tangent makes equal angle with the axis?

$\begin{array}{l}\mathrm{y}=\frac{2}{3}{\mathrm{x}}^3+\frac{1}{2}{\mathrm{x}}^2\\ \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2}{3}3{\mathrm{x}}^2+\frac{1}{2}2\mathrm{x}=2{\mathrm{x}}^2+\mathrm{x}\end{array}$

Since the tangent makes equal angles with the axes, $\frac{\mathrm{dy}}{\mathrm{dx}}=\pm 1$

$\begin{array}{l}\mathrm{or} 2{\mathrm{x}}^2+\mathrm{x}=\pm 1\\ \mathrm{or} 2{\mathrm{x}}^2+\mathrm{x}-1=0 \left(2{\mathrm{x}}^2+\mathrm{x}+1=0 \mathrm{has} \mathrm{no} \mathrm{real} \mathrm{roots}\right)\\ \mathrm{or} \left(2\mathrm{x}-1\right)\left(\mathrm{x}+1\right)=0\\ \mathrm{i}.\mathrm{e}, \mathrm{x}=\frac{1}{2} \mathrm{or} \mathrm{x}=-1\end{array}$