Step 1:
Equation of the tangent and normal to $x^2=9y$ at $(-3,1)$
The equation of the tangent is given by $xx_1=9\large\frac{(y+y_1)}{2}$
Where $(x_1,y_1)=(-3,1)$
Therefore the tangent is $-3x_1=\large\frac{9(y+1)}{2}$
(i.e) $6x+9y+9=0$
The above equation is divided by 3 we get
$2x+3y+3=0$
Step 2:
The normal is $\perp$ to the tangent .So its equation is of the form $3x-2y+k=0$
It passes through $(-3,1)$
$-9-2+k=0$
$\Rightarrow k=11$
The normal is $3x-2y+11=0$
Step 3:
The equation of tangent : $2x+3y+3=0$
The equation of normal : $3x-2y+11=0$