If y is directly proportional to x squared and y = 4 when x = 4, what is the value of y when x = 3?

Given that $y \propto x^2$, we can write this as:

$$y = kx^2$$

where $k$ is the constant of proportionality.

We are given that $y=4$ when $x=4$. Substitute these values into the equation:

$$4 = k(4^2)$$ $$4 = k(16)$$

To find $k$, divide both sides by 16:

$$k = \frac{4}{16}$$ $$k = \frac{1}{4}$$

Substitute the value of $k$ back into the equation $y = kx^2$:

$$y = \frac{1}{4}x^2$$

Substitute $x=3$ into the law:

$$y = \frac{1}{4}(3^2)$$ $$y = \frac{1}{4}(9)$$ $$y = \frac{9}{4}$$

$$y = 2\frac{1}{4}$$

Comparing this result with the given options:

(a) $4\frac{1}{2}$

(b) $3\frac{1}{16}$

(c) $2\frac{1}{4}$

(d) $1\frac{1}{2}$

The correct option is (c).

The final answer is \boxed{\text{2\frac{1}{4}}}.