Given $y=x^3+3x^2+5$

Now $\frac{dy}{dx}=3x^2+6x$

So at point $\left(x_1,y_1\right)$

${\left(\frac{dy}{dx}\right)}_{\mathrm{at} \left(x_1,y_1\right)}=3x_1^2+6x_1 ... \left(\mathrm{i}\right)$

Also slope of line in two point form will be $\frac{y_1-0}{x_1-0} ... \left(\mathrm{ii}\right)$

Also $\left(x_1,y_1\right)$ lie on curve

We get $y_1=x_1^3+3x_1^2+5 ... \left(\mathrm{iii}\right)$

From equation $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$ we get $3x_1^3+6x_1^2=y_1$ (as slope is equal)

Now $y_1=3x_1^3+6x_1^2=x_1^3+3x_1^2+5$

$\Rightarrow 2y_1=3x_1^2+\frac{15}{2}$

i.e. $2y=3x_{}^2+\frac{15}{2}$ does not intersect the curve $\frac{x}{3}-y^2=2$