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Let's tackle the problem step by step:

### Part 1: Solving for \( x \) in the Equation
The given equation is:
\[
x^2 + \frac{9}{x} = 10
\]

To solve for \( x \):
1. Rearrange the equation:
   \[
   x^2 + \frac{9}{x} - 10 = 0
   \]
2. Multiply through by \( x \) (assuming \( x \neq 0 \)) to eliminate the fraction:
   \[
   x^3 - 10x + 9 = 0
   \]

This is a cubic equation. We can solve it by testing possible rational roots using the Rational Root Theorem. After testing, we find that \( x = 1 \) is a solution. 

3. Factor \( x - 1 \) out of the polynomial using synthetic division:
   - \( x^3 - 10x + 9 = (x - 1)(x^2 + x - 9) \)
   
4. Now solve \( x^2 + x - 9 = 0 \) using the quadratic formula:
   \[
   x = \frac{-1 \pm \sqrt{1 + 36}}{2} = \frac{-1 \pm \sqrt{37}}{2}
   \]

Thus, the solutions are:
\[
x = 1, \quad x = \frac{-1 + \sqrt{37}}{2}, \quad x = \frac{-1 - \sqrt{37}}{2}
\]

### Part 2: Solving the Inequality
The inequality is:
\[
x^3 - 3x > 0
\]

Factor the inequality:
\[
x(x^2 - 3) > 0
\]

Factor further to:
\[
x(x - \sqrt{3})(x + \sqrt{3}) > 0
\]

Now, use the sign analysis on the intervals determined by the roots \( x = 0, x = \sqrt{3}, \) and \( x = -\sqrt{3} \):
1. For \( x \in (-\infty, -\sqrt{3}) \), \( x(x - \sqrt{3})(x + \sqrt{3}) < 0 \).
2. For \( x \in (-\sqrt{3}, 0) \), \( x(x - \sqrt{3})(x + \sqrt{3}) > 0 \).
3. For \( x \in (0, \sqrt{3}) \), \( x(x - \sqrt{3})(x + \sqrt{3}) < 0 \).
4. For \( x \in (\sqrt{3}, \infty) \), \( x(x - \sqrt{3})(x + \sqrt{3}) > 0 \).

The inequality \( x^3 - 3x > 0 \) holds for \( x \in (-\sqrt{3}, 0) \cup (\sqrt{3}, \infty) \).

Therefore, the solution to both the equation and the inequality is:
\[
x = \frac{-1 + \sqrt{37}}{2} \quad \text{or} \quad x = 1 \quad \text{as these satisfy both conditions.}
\]

### Part 3: Showing \( y = x^3 - 3x \) is Increasing at \( x = 1 \)
To check if \( y \) is increasing at \( x = 1 \):
1. Find \( y' = \frac{d}{dx}(x^3 - 3x) = 3x^2 - 3 \).
2. Evaluate \( y' \) at \( x = 1 \):
   \[
   y'(1) = 3(1)^2 - 3 = 3 - 3 = 0
   \]

Since \( y' = 0 \) at \( x = 1 \), the function is not increasing at this point in the usual sense; it is a critical point, specifically a point of inflection.

Would you like more details on any specific step, or have further questions?

### Related Questions:
1. How can we solve a general cubic equation using the Rational Root Theorem?
2. What methods are available for solving polynomial inequalities?
3. How can we determine the intervals of increase and decrease for functions?
4. Why is the point \( x = 1 \) a point of inflection for \( y = x^3 - 3x \)?
5. How does the quadratic formula work, and when is it applicable?

#### Tip:
When dealing with inequalities involving polynomials, sketching the graph or performing a sign analysis can help visualize solution intervals quickly.

Consider the equation x^2 + 9/x = 10. Solve for x such that the inequality x^3 - 3x > 0 is also satisfied. Additionally, show that y = x^3 - 3x is increasing at x = 1.

Learn how to solve the equation x^2 + 9/x = 10 and find the values of x that also satisfy the inequality x^3 - 3x > 0. Additionally, see how to show that y = x^3 - 3x is increasing at x = 1 using calculus. Suitable for advanced high school algebra students.

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