To solve the given equation $x^2 + \frac{9}{x^2} = 10$, let's start by multiplying both sides of the equation by $x^2$ to get rid of the fraction. This gives us $x^4 + 9 = 10x^2$.

Rearranging the equation, we get $x^4 - 10x^2 + 9 = 0$. This is a quadratic equation in terms of $x^2$, so we can use the quadratic formula to solve for $x^2$.

The quadratic formula is given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. In this case, $a = 1$, $b = -10$, and $c = 9$. Plugging these values into the formula, we get $x^2 = \frac{10 \pm \sqrt{(-10)^2 - 4(1)(9)}}{2(1)}$.

Simplifying the expression, we get $x^2 = \frac{10 \pm \sqrt{100 - 36}}{2} = \frac{10 \pm \sqrt{64}}{2} = \frac{10 \pm 8}{2}$.

This gives us two possible values for $x^2$: $x^2 = \frac{10 + 8}{2} = 9$ and $x^2 = \frac{10 - 8}{2} = 1$.

Taking the square root of both sides, we get $x = \pm 3$ and $x = \pm 1$.

Now, we need to check which of these solutions satisfy the inequality $x^3 - 3x > 0$.

We can start by plugging in $x = 3$ into the inequality: $3^3 - 3(3) = 27 - 9 = 18 > 0$. So, $x = 3$ satisfies the inequality.

Next, we plug in $x = -3$: $(-3)^3 - 3(-3) = -27 + 9 = -18 < 0$. So, $x = -3$ does not satisfy the inequality.

Now, we plug in $x = 1$: $1^3 - 3(1) = 1 - 3 = -2 < 0$. So, $x = 1$ does not satisfy the inequality.

Finally, we plug in $x = -1$: $(-1)^3 - 3(-1) = -1 + 3 = 2 > 0$. So, $x = -1$ satisfies the inequality.

Therefore, the solutions to the equation that also satisfy the inequality are $x = 3$ and $x = -1$.

Now, let's show that $y = x^3 - 3x$ is increasing at $x = 1$. To do this, we need to find the derivative of $y$ with respect to $x$ and evaluate it at $x = 1$.

Using the power rule of differentiation, we get $\frac{dy}{dx} = 3x^2 - 3$.

Evaluating the derivative at $x = 1$, we get $\frac{dy}{dx}(1) = 3(1)^2 - 3 = 3 - 3 = 0$.

At $\frac{dy}{dx}(1) = 0$, $y$ is neither increasing nor decreasing. However, we can show that the function is increasing by looking at the behavior of the derivative around $x = 1$. We can do this by looking at the derivative at $x = 1 + \epsilon$ for some small $\epsilon > 0$.