Step 1:

$y=4x^3-3x+5$

On differentiating with respect to $x$

$\large\frac{dy}{dx}=$$12x-3$-------(1)

It is $\perp$ to the line $9y+x+3=0$

The slope of the line is $-(\large\frac{1}{9})$

$\therefore$ The slope of the tangent is $m_1(\large\frac{-1}{9})=$$-1$

$\Rightarrow m_1=9$

$\therefore \large\frac{dy}{dx}$$=9$-----(2)

Equating equ(1) and equ(2)

$12x-3=9$

$12x=12$

$x=1$

If $x=1$,$y=4(1)^3-3(1)+5,y=6$

$\therefore x=1$ and $y=6$

Hence the point of contact of the tangent to the curve is $(x_1,y_1)=(1,6)$

Step 2:

Equation of the tangent is $y-y_1=m(x-x_1)$

$\Rightarrow y-6=9(x-1)$

$\Rightarrow y-6=9x-9$

$9x-y=3$

This is the required equation.