We have, $y=-x^3+3 x^2+9 x-27$
$\therefore \frac{d y}{d x}=-3 x^2+6 x+9=$ Slope of tangent to the curve
Now, $\frac{d^2 y}{d x^2}=-6 x+6$
For maxzimum $\frac{d y}{d x}\left(\frac{d y}{d x}\right)=0$
$$
\Rightarrow \mathrm{x}=\frac{-6}{-6}=1 \quad \therefore \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\right)=-6 < 0
$$
So, the slope of tangent to the curve is maximum, when $\mathrm{x}=1$,
For $\mathrm{x}=1$,
$$
\left(\frac{d y}{d x}\right)_{(x=1)}=-3.1^2+6.1+9=12,
$$
which is maximum slope.
Also, for $\mathrm{x}=1$,
$$
y=-1^3+3.1^2+9.1-27=-16
$$
So, the required point is $(1,-16)$.