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We are tasked with finding the equations of all tangent lines to the curve \( -3 + x^2 = xy \) when \( x = 3 \).

### Step 1: Implicit Differentiation
The given equation is:

\[
-3 + x^2 = xy
\]

We'll differentiate both sides with respect to \( x \), keeping in mind that \( y \) is a function of \( x \) (so we must apply the product and chain rules).

The derivative of the left-hand side:

\[
\frac{d}{dx} \left( -3 + x^2 \right) = 2x
\]

Now for the right-hand side, using the product rule on \( xy \):

\[
\frac{d}{dx}(xy) = x \frac{dy}{dx} + y
\]

Thus, the implicit differentiation gives:

\[
2x = x \frac{dy}{dx} + y
\]

### Step 2: Solve for \( \frac{dy}{dx} \)
Now, isolate \( \frac{dy}{dx} \):

\[
2x - y = x \frac{dy}{dx}
\]

\[
\frac{dy}{dx} = \frac{2x - y}{x}
\]

### Step 3: Find \( y \) when \( x = 3 \)
Substitute \( x = 3 \) into the original equation to find \( y \):

\[
-3 + 3^2 = 3y
\]

\[
-3 + 9 = 3y
\]

\[
6 = 3y
\]

\[
y = 2
\]

### Step 4: Find the Slope of the Tangent Line
Substitute \( x = 3 \) and \( y = 2 \) into the derivative formula:

\[
\frac{dy}{dx} = \frac{2(3) - 2}{3} = \frac{6 - 2}{3} = \frac{4}{3}
\]

Thus, the slope of the tangent line at \( x = 3 \) is \( \frac{4}{3} \).

### Step 5: Equation of the Tangent Line
We now use the point-slope form of the equation of a line:

\[
y - y_1 = m(x - x_1)
\]

where \( m = \frac{4}{3} \), \( x_1 = 3 \), and \( y_1 = 2 \). Substituting these values:

\[
y - 2 = \frac{4}{3}(x - 3)
\]

Simplifying:

\[
y - 2 = \frac{4}{3}x - 4
\]

\[
y = \frac{4}{3}x - 2
\]

Thus, the equation of the tangent line at \( x = 3 \) is:

\[
y = \frac{4}{3}x - 2
\]

Would you like more details, or do you have any questions?

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### 5 Related Questions:
1. How do you handle implicit differentiation with multiple variables?
2. What is the significance of the point-slope form in finding tangent lines?
3. How do you find the tangent line to a curve if the equation is not easily solvable for \( y \)?
4. Can there be more than one tangent line at a given \( x \) value for different types of curves?
5. How does implicit differentiation differ from explicit differentiation in complex functions?

### Tip:
Always double-check your derivative rules when working with implicit differentiation, especially with products of functions like \( xy \).

If -3 + x^2 = xy then find the equations of all tangent lines to the curve when x = 3.

Explore how to find the equations of tangent lines to the curve defined by -3 + x^2 = xy at x = 3. This detailed guide applies implicit differentiation, uses the product rule, and the point-slope form of a line to derive the tangent lines. Suitable for college students studying calculus.

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