Find y' for x²+xy+y²=6
Find $y'$ for $x^{2}+xy+y^{2}=6$.
Explanation of Method
A tilted ellipse, differentiated without solving for y.
Step 1: Differentiate both sides with respect to $x$
The middle term is a product of two functions of $x$, so it needs the product rule; the last needs the chain rule: $2x + \bigl(y + xy'\bigr) + 2y\,y' = 0$.
Step 2: Collect the terms containing $y'$
\[ y'\bigl(x+2y\bigr) = -2x-y. \]
Step 3: Solve
\[ y' = -\frac{2x+y}{x+2y}. \]
Step 4: Verification
Take $x = 1$. Then $1+y+y^{2}=6$, so $y^{2}+y-5=0$ and $y = \dfrac{-1+\sqrt{21}}{2} \approx 1.79129$ on the upper branch.
\[y'(1) = -\frac{2+1.79129}{1+3.58258} = -\frac{3.79129}{4.58258} \approx -0.82733\]
\[\text{Numerically: } x = 1\pm 10^{-5} \;\Longrightarrow\; \frac{\Delta y}{\Delta x} \approx -0.82733 \;\;\checkmark\]
\[x = 0:\quad y^{2} = 6,\ y = \sqrt6; \qquad y' = -\frac{\sqrt6}{2\sqrt6} = -\frac12 \;\;\checkmark\]
Step 5: Horizontal and vertical tangents
The numerator vanishes when $y = -2x$; substituting into the curve gives $x^{2}-2x^{2}+4x^{2}=6$, i.e. $x = \pm\sqrt2$ — two points with horizontal tangents. The denominator vanishes when $x = -2y$, giving $y = \pm\sqrt2$ and two vertical tangents.
Result
Final answer:
\[\boxed{\,y' = -\dfrac{2x+y}{x+2y}, \qquad x+2y \neq 0\,}\]
Remark
Remark. The quadratic $x^{2}+xy+y^{2}$ has discriminant $1-4 = -3 \lt 0$, so the curve is a genuine ellipse, tilted $45^{\circ}$ by the cross term. The answer has the pleasing symmetry $y'(x,y) = 1/y'(y,x)$, reflecting the fact that swapping $x$ and $y$ maps the curve to itself.