To solve the equation

$x^2 + \frac{9}{x^2} = 10,$

we can first eliminate the fraction by multiplying through by $x^2$ (assuming $x \neq 0$):

$x^4 - 10x^2 + 9 = 0.$

Let $u = x^2$. The equation then becomes a quadratic:

$u^2 - 10u + 9 = 0.$

We can solve this using the quadratic formula:

$u = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 9}}{2 \cdot 1} = \frac{10 \pm \sqrt{100 - 36}}{2} = \frac{10 \pm \sqrt{64}}{2} = \frac{10 \pm 8}{2}.$

This gives us:

$u = \frac{18}{2} = 9 \quad \text{or} \quad u = \frac{2}{2} = 1.$

Recalling that $u = x^2$, we have:

$x^2 = 9 \quad \Rightarrow \quad x = 3 \text{ or } x = -3,$ $x^2 = 1 \quad \Rightarrow \quad x = 1 \text{ or } x = -1.$

Next, we will consider the inequality $x^3 - 3x > 0$. We can factor this expression:

$x^3 - 3x = x(x^2 - 3).$

Setting $x(x^2 - 3) > 0$, we find the critical points:

1. $x = 0$,
2. $x^2 - 3 = 0 \quad \Rightarrow \quad x = \sqrt{3}, \, -\sqrt{3}$.

Now we will analyze the sign of $x(x^2 - 3)$ in the intervals defined by these critical points:

• For $x < -\sqrt{3}$, $x < 0$ and $x^2 - 3 > 0$ (both factors negative): positive.
• For $-\sqrt{3} < x < 0$, $x < 0$ and $x^2 - 3 < 0$ (one factor negative, one positive): negative.
• For $0 < x < \sqrt{3}$, $x > 0$ and $x^2 - 3 < 0$ (one factor positive, one negative): negative.
• For $x > \sqrt{3}$, $x > 0$ and $x^2 - 3 > 0$ (both factors positive): positive.

Thus, the solution for $x^3 - 3x > 0$ is

$x < -\sqrt{3} \quad \text{or} \quad x > \sqrt{3}.$

Now we compare the solutions $x = 3$, $x = -3$, $x = 1$, and $x = -1$ with the inequality:

• $x = 3$ satisfies $x > \sqrt{3}$ (valid).
• $x = -3$ satisfies $x < -\sqrt{3}$ (valid).
• $x = 1$ does not satisfy $x > \sqrt{3}$ or $x < -\sqrt{3}$ (not valid).
• $x = -1$ does not satisfy $x > \sqrt{3}$ or $x < -\sqrt{3}$ (not valid).

Therefore, the valid solutions for the equation that also satisfy the inequality are

$x = 3 \quad \text{and} \quad x = -3.$

Next, we check if $y = x^3 - 3x$ is increasing at $x = 1$. To do this, we compute the derivative:

$y' = \frac{d}{dx}(x^3 - 3x) = 3x^2 - 3.$

Evaluating the derivative at $x = 1$:

$y'(1) = 3(1)^2 - 3 = 3 - 3 = 0.$

Since $y'(1) = 0$, the function $y = x^3 - 3x$ is neither increasing nor decreasing at $x = 1$. To determine the behavior near $x = 1$, we check the second derivative:

$y'' = \frac{d^2}{dx^2}(x^3 - 3x) = 6x.$

At $x = 1$:

$y''(1) = 6(1) = 6 > 0.$

Thus, $y$ has a local minimum at $x = 1$, indicating that $y$ is increasing for $x > 1$ and decreasing for $x < 1$.

In conclusion, the valid solutions are:

$\boxed{3 \text{ and } -3}$

and the function $y = x^3 - 3x$ is at a local minimum at $x = 1$. It does not increase at this point, but is increasing for $x > 1$.

Would you like further details on any part of this solution? Here are some related questions to consider:

1. What are the graphical representations of $y = x^3 - 3x$?
2. How does the function $y$ behave for $x < -\sqrt{3}$?
3. What happens to the function's behavior at other critical points?
4. Can we derive an expression for the points of inflection of $y$?
5. How would the inequality change if it was $x^3 - 3x \leq 0$?

And as a tip: Always check the behavior of a function by analyzing critical points and testing intervals to understand where it increases or decreases!