The equation of the given curve is

$\begin{array}{l}

y=4 x^{3}-2 x^{5} \\

\frac{\mathrm{dy}}{\mathrm{dx}}=12 x^{2}-10 x^{4}

\end{array}$

Therefore, the slope of the tangent at point $(x, y)$ is $12 x^{2}-10 x^{4}$. The equation of the tangent at $(x, y)$ is given by

$\mathrm{Y}-y=\left(12 x^{2}-10 x^{4}\right)(\mathrm{X}-x) \ldots \text { (i) }$

When, the tangent passes through the origin $(0,0)$, then $\mathrm{X}=\mathrm{Y}=0$ Therefore, eq. (i) reduce to

$\begin{array}{l}

-y=\left(12 x^{2}-10 x^{4}\right)(-x) \\

\Rightarrow y=12 x^{3}-10 x^{5}

\end{array}$

Also, we have $y=4 x^{3}-2 x^{5}$

$\begin{array}{l}

\therefore 12 x^{3}-10 x^{5} \\

=4 x^{3}-2 x^{5} \\

\Rightarrow 8 x^{5}-8 x^{3}=0 \\

\Rightarrow x^{5}-x^{3}=0 \\

\Rightarrow x^{3}\left(x^{2}-1\right)=0 \\

\Rightarrow x=0, \pm 1

\end{array}$

When, $x=0$

$y=4(0)^{3}-2(0)^{5}=0$

When, $x=1$,

$y=4(1)^{3}-2(1)^{5}=2$

When, $x=-1$,

$y=4(-1)^{3}-2(-1)^{5}=-2$

Hence, the require points are $(0,0),(1,2)$ and $(-1,-2)$.