Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 25

Find an equation of the tangent to the curve $x^2 + xy + y^2 = 3$ at the point $(1,1)$ using Implicit Differentiation.

If $y'=m\text{ (slope)}$ then
$\displaystyle \frac{d}{dx} (x^2) + \frac{d}{dx} (xy) + \frac{d}{dx} (y^2) = \frac{d}{dx} (3)$

$
\begin{equation}
\begin{aligned}
\frac{d}{dx} (x^2) + \left[ (x) \frac{d}{dx} (y) + (y) \frac{d}{dx} (x) \right] + \frac{d}{dx} (y^2) &= \frac{d}{dx} (3)\\
\\
2x + (x) \frac{dy}{dx} + (y)(1) + 2y \frac{dy}{dx} &= 0\\
\\
2x + xy' + y + 2yy' &= 0\\
\\
xy'+2yy' &= -2x-y\\
\\
y'(x+2y) &= -2x-y\\
\\
\frac{y'\cancel{(x+2y)}}{\cancel{x+2y}} &= \frac{-2x-y}{x+2y}\\
\\
y' = m &= \frac{-2x-y}{x+2y}
\end{aligned}
\end{equation}
$


For $x =1 $ and $y = 1 $ , we have

$
\begin{equation}
\begin{aligned}
m &= \frac{-2(1)-1}{1+2(1)}\\
\\
m &= \frac{-2-1}{1+2}\\
\\
m &= \frac{-3}{3}\\
\\
m &= -1
\end{aligned}
\end{equation}
$


Using point slope form,



$
\begin{equation}
\begin{aligned}
y-y_1 &= m (x-x_1)\\
\\
y - 1 &= -1 (x-1)\\
\\
y - 1 &= -x + 1\\
\\
y &= -x + 1 +1 \\
\\
y & = -x + 2 && \text{Equation of the tangent at } (1,1)
\end{aligned}
\end{equation}
$