The point on the curve

The curve $y=6 x-x^{2}$ has a point at which the tangent to the curve is inclined at to $\frac{\pi}{4}$ the line $x+y=0$.

Differentiating w.r.t. $x$,

$\frac{\mathrm{dy}}{\mathrm{dx}}=6-2 \mathrm{x}=\mathrm{m}_{1}$ and $\frac{\mathrm{dy}}{\mathrm{dx}}=-1=\mathrm{m}_{2}$

$\tan \theta=\left|\frac{\mathrm{m}_{1}-\mathrm{m}_{2}}{1+\mathrm{m}_{1} \mathrm{~m}_{2}}\right|$

$\Rightarrow \tan \frac{\pi}{4}=\left|\frac{6-2 \mathrm{x}+1}{1+2 \mathrm{x}-6}\right|$

On solving we get $x=3$

Thus $y=9$

Hence, option B is correct.