\begin{align*}
5x^2-6xy(x)+2(y(x))^2=1 \amp \implies \frac{d}{dx}(5x^2-6xy(x)+2(y(x))^2)=\frac{d}{dx}(1) \\
\amp \implies 10x-6(x\, y(x))'+4y(x)y'(x)=0\\
\amp \implies 10x-6(y(x)+xy'(x))+4y(x)y'(x)=0\\
\amp \implies 10x-6y(x)-6xy'(x)+4y(x)y'(x)=0\\
\amp \implies y'(x)=\frac{6y(x)-10x}{4y(x)-6x}\text{.}
\end{align*}
We look for point \(P=(x,y)\) where \(y'(x)=0\text{,}\) making the tangent line horizontal. According to our derivation above, for this to be the case we need need \(6y-10x=0\text{,}\) or equivalently, \(y=\frac{5}{3}x\text{.}\) Substituting this back in to the defining equation of \(\mathcal{C}\text{,}\) we see that
\begin{equation*}
5x^2-6x(5x/3)+2(5x/3)^2=1\text{,}
\end{equation*}
or
\begin{equation*}
\frac{5}{9}x^2=1\text{.}
\end{equation*}
We conclude that \(x=\pm 3/\sqrt{5}\text{.}\) Using \(y=\frac{5}{3}x\text{,}\) we see that the points
\begin{align*}
P \amp =(3/\sqrt{5},5/\sqrt{5})\amp Q\amp=(-3/\sqrt{5},-5/\sqrt{5})
\end{align*}
are the two points where the tangent lines to \(\mathcal{C}\) are horizontal. Below you find a graph of \(\mathcal{C}\text{,}\) the points \(P\) and \(Q\text{,}\) and their horizontal tangent lines.
