Step 1:
Equation of the rectangular hyperbola is $xy=12$.
Equation of the tangent to $xy=12$ at $(x_1,y_1)$ is $\large\frac{1}{2}$$(xy_1+yx_1)$$=12$
Here $(x_1,y_1)=(3,4)$
Therefore the tangent is $4x+3y=24$
Step 2:
The normal is $\perp$ to the tangent and the equation of the form $3x-4y=k$.It passes through $(3,4)$
Therefore $9-16=k$
$\Rightarrow k=-7$
The normal is $3x-4y=-7$
$\Rightarrow 3x-4y+7=0$
Step 3:
Equation of tangent : $4x+3y=24$
Equation of normal : $3x-4y+7=0$