Hello, sweetG!
a) Find the equation of the normal to $\displaystyle y \:= \:\frac{1}{3}x^2$ where $\displaystyle x = 1$
When $\displaystyle x = 1,\;y = \frac{1}{3}$
We have: .$\displaystyle y' \:=\:\frac{2}{3}x$
When $\displaystyle x = 1$, the slope of the tangent is: .$\displaystyle m \:=\:\frac{2}{3}$
. . Hence, the slope of the normal is: $\displaystyle \frac{3}{2}$
The equation of the normal is: .$\displaystyle y  \frac{1}{3}\:=\:\frac{3}{2}(x1) \quad\Rightarrow\quad \boxed{y \:=\:\frac{3}{2}x + \frac{11}{6}}$
b) This normal cuts the coordinate axis at P and Q.
Given that P and Q are adjacent vertices of a rhombus,
and that the diagonals of this rhombus intersect at the origin,
determine the area of this rhombus.
The intercept of the normal are: .$\displaystyle P\left(\frac{11}{9},\:0\right)\,\text{ and }\,Q\left(0,\:\frac{11}{6}\right)$
If the diagonals intersect at the origin, the rhombus looks like this: Code:

Q o (0,11/6)
*  *
*  *
R *  * P
   o      +      o   
(11/9,0) *  * (11/9,0)
*  *
*  *
S o (0,11/6)

The area of a rhomus is onehalf the product of its diagonals.
The horizontal diagonal has length: .$\displaystyle 2 \times \frac{11}{9} \:=\:\frac{22}{9}$
The vertical diagonal has length: .$\displaystyle 2 \times \frac{11}{6} \:=\:\frac{11}{3}$
The area is: .$\displaystyle \frac{1}{2} \times \frac{22}{9} \times \frac{11}{3} \;=\;\boxed{\frac{121}{27}\text{ units}^2} $
*