To find the equation of the tangent line using implicit differentiation, we'll follow these steps:

Step 1: Differentiate the equation implicitly

The given equation is: $x^2 - xy - y^2 = 1$ We will differentiate both sides with respect to $x$, remembering that $y$ is a function of $x$, so whenever we differentiate terms involving $y$, we must use the chain rule.

$\frac{d}{dx}(x^2) - \frac{d}{dx}(xy) - \frac{d}{dx}(y^2) = \frac{d}{dx}(1)$

Differentiating each term:

1. $\frac{d}{dx}(x^2) = 2x$
2. $\frac{d}{dx}(xy)$: We apply the product rule here: $\frac{d}{dx}(xy) = \frac{d}{dx}(x) \cdot y + x \cdot \frac{d}{dx}(y)$, which simplifies to $y + x \frac{dy}{dx}$
3. $\frac{d}{dx}(y^2)$: Using the chain rule: $\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}$
4. The derivative of a constant is 0: $\frac{d}{dx}(1) = 0$

So, the equation becomes: $2x - (y + x \frac{dy}{dx}) - 2y \frac{dy}{dx} = 0$

Step 2: Solve for $\frac{dy}{dx}$

Now, we need to solve for $\frac{dy}{dx}$, the slope of the tangent line.

$2x - y - x \frac{dy}{dx} - 2y \frac{dy}{dx} = 0$

Group the terms with $\frac{dy}{dx}$ on one side: $2x - y = x \frac{dy}{dx} + 2y \frac{dy}{dx}$

Factor out $\frac{dy}{dx}$ on the right-hand side: $2x - y = (x + 2y) \frac{dy}{dx}$

Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{2x - y}{x + 2y}$

Step 3: Substitute the given point $(2, 1)$

Substitute $x = 2$ and $y = 1$ into the derivative to find the slope at that point: $\frac{dy}{dx} = \frac{2(2) - 1}{2 + 2(1)} = \frac{4 - 1}{2 + 2} = \frac{3}{4}$

So, the slope of the tangent line at the point $(2, 1)$ is $\frac{3}{4}$.

Step 4: Use the point-slope form to find the equation of the tangent line

The point-slope form of a line is: $y - y_1 = m(x - x_1)$ where $m$ is the slope and $(x_1, y_1)$ is the point on the line.

Substituting $m = \frac{3}{4}$, $x_1 = 2$, and $y_1 = 1$ into the equation: $y - 1 = \frac{3}{4}(x - 2)$

Simplifying this: $y - 1 = \frac{3}{4}x - \frac{3}{2}$ $y = \frac{3}{4}x - \frac{3}{2} + 1$ $y = \frac{3}{4}x - \frac{1}{2}$

Thus, the equation of the tangent line is: $y = \frac{3}{4}x - \frac{1}{2}$

Summary:

The equation of the tangent line to the curve $x^2 - xy - y^2 = 1$ at the point $(2, 1)$ is: $y = \frac{3}{4}x - \frac{1}{2}$

Would you like further details or have any questions about the steps? Here are some related questions you might find useful:

1. How would the solution change if the point was different, say $(1, 3)$?
2. How do you check if the equation of the tangent line is correct by using the original equation?
3. What would the equation of the tangent line be for a curve defined by an exponential function?
4. Can you find the second derivative from the implicit function?
5. How would the slope change if the equation of the curve had a different form, such as $x^3 - xy + y^2 = 1$?

Tip: When using implicit differentiation, always remember to apply the chain rule to terms involving $y$ since $y$ is treated as a function of $x$.