(b)$y=\frac{1}{2} x+\frac{7}{2}$

(a)
$x^{\frac{1}{2}}+y^{\frac{1}{2}} =3-(1)$

$\frac{\mathrm{d}}{\mathrm{d} x}\left(x^{\prime}\right)+\frac{\mathrm{d}}{\mathrm{d} x}\left(y^{\frac{1}{2}}\right) =\frac{\mathrm{d}}{\mathrm{d} x}(3)$

$\frac{1}{2} x^{-\frac{1}{2}}+\frac{1}{2} y^{-\frac{1}{2}} \frac{\mathrm{d} y}{\mathrm{~d} x} =0 $

$\frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}} \frac{\mathrm{d} y}{\mathrm{~d} x} =0$
$\frac{1}{2 \sqrt{y}} \frac{\mathrm{d} y}{\mathrm{~d} x} =-\frac{1}{2 \sqrt{x}} $
$\frac{\mathrm{d} y}{\mathrm{~d} x} =\left(-\frac{1}{2 \sqrt{x}}\right)(2 \sqrt{y})$
$=-\frac{\sqrt{y}}{\sqrt{x}}$

$\therefore \frac{\mathrm{d} y}{\mathrm{~d} x}=-\left(\frac{y}{x}\right)^\frac{1}{2}$

(b) When $x=1$,
(1): $1+y^{\frac{1}{2}}=3$
$y^1=2 $
$y=2^2$
$y=4$

When $x=1, y=4, \frac{\mathrm{d} y}{\mathrm{~d} x}=-\left(\frac{4}{1}\right)^{\frac{1}{2}}=-\sqrt{4}=-2$
Gradient of normal $=\frac{1}{2}$
Equation of normal to curve at $(1,4)$ :
$y-4 =\frac{1}{2}(x-1)$

$y-4 =\frac{1}{2} x-\frac{1}{2} $
$y =\frac{1}{2} x-\frac{1}{2}+4 $
$y =\frac{1}{2} x+\frac{7}{2}$