If y varies inversely as the square of x, and the difference between the values of y when x=6 and x=10 is 16, find x when y=4.

The problem states that $y$ varies inversely as the square of $x$. This can be written as:

$$y = \frac{k}{x^2}$$

where $k$ is the constant of proportionality.

We are given that the difference between the values of $y$ when $x=6$ and $x=10$ is 16.
When $x=6$, $y_1 = \frac{k}{6^2} = \frac{k}{36}$.
When $x=10$, $y_2 = \frac{k}{10^2} = \frac{k}{100}$.
Since $y$ varies inversely with $x^2$, a smaller $x$ value will result in a larger $y$ value. Thus, $y_1 > y_2$.
The difference is $y_1 - y_2 = 16$.

$$\frac{k}{36} - \frac{k}{100} = 16$$

To solve for $k$, find a common denominator for 36 and 100. The least common multiple of 36 and 100 is 900.

$$\frac{25k}{900} - \frac{9k}{900} = 16$$ $$\frac{25k - 9k}{900} = 16$$ $$\frac{16k}{900} = 16$$

Multiply both sides by 900:

$$16k = 16 \times 900$$

Divide both sides by 16:

$$k = 900$$

Substitute the value of $k$ back into the variation equation from Step 1:

$$y = \frac{900}{x^2}$$

Substitute $y=4$ into the law found in Step 3:

$$4 = \frac{900}{x^2}$$

Multiply both sides by $x^2$:

$$4x^2 = 900$$

Divide both sides by 4:

$$x^2 = \frac{900}{4}$$ $$x^2 = 225$$

Take the square root of both sides:

$$x = \pm\sqrt{225}$$ $$x = \pm 15$$

The law connecting $x$ and $y$ is $\boxed{y = \frac{900}{x^2}}$.
When $y=4$, $x = \boxed{\pm 15}$.