Explanation

(b)
Given curve is $x^{2}y = x^{2} - 3x + 6$ …(i)
At $x = 3,\ \ \ \ 3^{2}(y) = 3^{2} - 3(3) + 6$
$\Rightarrow \ \ y = \frac{2}{3}$
On differentiating Eq. (i) w.r.t. $x$, we get
$2xy + x^{2}\frac{dy}{dx} = 2x - 3$
$\Rightarrow \ \ \frac{dy}{dx} = \frac{2x - 3 - 2xy}{x^{2}}$
$\Rightarrow \ \ \ \ \left( \frac{dy}{dx} \right)_{\left( 3,\frac{2}{3} \right)} = \frac{6 - 3 - 2 \times 3 \times \frac{2}{3}}{3^{2}} = - \frac{1}{3^{2}}$
$\therefore$ Equating of normal is
$y - \frac{2}{3} = 3^{2}(x - 3)$
$\Rightarrow \ \ \ \ \ 27x - 3y = 79$