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La respuesta: los puntos donde es perpendicular son (0,-1), (1,1) y (-1,-1)

The problem asks us to find the points on the graph of the equation:

\[
x^2 + \frac{y}{x} = 2
\]

where the tangent line is perpendicular to the line given by:

\[
2x - 2y = 5.
\]

### Step-by-step approach:

1. **Find the slope of the given line**:  
   We rewrite the equation \(2x - 2y = 5\) in slope-intercept form \(y = mx + b\):
   \[
   2x - 2y = 5 \quad \Rightarrow \quad y = x - \frac{5}{2}.
   \]
   The slope of the line is \(m = 1\).

2. **Perpendicular slope**:  
   For the tangent to be perpendicular to this line, the slope of the tangent should be the negative reciprocal of 1, which is \(-1\).

3. **Implicit differentiation of the curve**:  
   Now, let's differentiate the given equation \(x^2 + \frac{y}{x} = 2\) implicitly with respect to \(x\).

   - The derivative of \(x^2\) is \(2x\).
   - The derivative of \(\frac{y}{x}\) is obtained by applying the quotient rule:
     \[
     \frac{d}{dx} \left(\frac{y}{x}\right) = \frac{x \frac{dy}{dx} - y}{x^2}.
     \]

   So, implicit differentiation of the entire equation gives:
   \[
   2x + \frac{x \frac{dy}{dx} - y}{x^2} = 0.
   \]

4. **Solve for \(\frac{dy}{dx}\)** (the slope of the tangent line):  
   Multiply through by \(x^2\) to eliminate the denominator:
   \[
   2x^3 + x \frac{dy}{dx} - y = 0.
   \]
   Now isolate \(\frac{dy}{dx}\):
   \[
   x \frac{dy}{dx} = y - 2x^3 \quad \Rightarrow \quad \frac{dy}{dx} = \frac{y - 2x^3}{x}.
   \]

5. **Set the slope equal to \(-1\)**:  
   Since we want the tangent line to be perpendicular to the given line, we set \(\frac{dy}{dx} = -1\):
   \[
   \frac{y - 2x^3}{x} = -1.
   \]
   Multiply through by \(x\):
   \[
   y - 2x^3 = -x \quad \Rightarrow \quad y = 2x^3 - x.
   \]

6. **Substitute into the original equation**:  
   Now substitute \(y = 2x^3 - x\) into the original equation \(x^2 + \frac{y}{x} = 2\):
   \[
   x^2 + \frac{2x^3 - x}{x} = 2.
   \]
   Simplify:
   \[
   x^2 + 2x^2 - 1 = 2 \quad \Rightarrow \quad 3x^2 - 1 = 2 \quad \Rightarrow \quad 3x^2 = 3 \quad \Rightarrow \quad x^2 = 1.
   \]
   So, \(x = 1\) or \(x = -1\).

7. **Find the corresponding \(y\) values**:  
   - If \(x = 1\), then \(y = 2(1)^3 - 1 = 2 - 1 = 1\).
   - If \(x = -1\), then \(y = 2(-1)^3 - (-1) = -2 + 1 = -1\).

8. **Solution points**:  
   Therefore, the points where the tangent line is perpendicular to the given line are \((1, 1)\) and \((-1, -1)\).

Additionally, let's check if there are more points by analyzing the form of the equation, but the possible points that satisfy both conditions are \((1, 1)\) and \((-1, -1)\).

---

Would you like more details on any specific part of the solution? Here are five related questions to explore:

1. How is implicit differentiation used to find the slope of a tangent line?
2. Why is the slope of a perpendicular line the negative reciprocal?
3. Can you show more examples of solving implicit differentiation problems?
4. What happens if we change the equation of the line to a different slope?
5. What other methods can be used to find points of tangency?

**Tip**: When solving problems involving perpendicular slopes, always remember that the product of their slopes is \(-1\).

Halle el o los puntos sobre la gráfica de ecuación x^2 + y/x = 2 donde la recta tangente es perpendicular a la recta 2x - 2y = 5.

Learn how to solve a problem involving implicit differentiation and perpendicular slopes. The task is to find the points on the curve x^2 + y/x = 2 where the tangent is perpendicular to the line 2x - 2y = 5. A step-by-step solution is provided, using algebra and calculus principles.

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