Answer: At \((3,1)\), \(\frac{dy}{dx}=0\). Also \(\frac{d^2y}{dx^2}=-\frac1{16}\), so the turning point is a local maximum.

Differentiate \(x^2-6xy+25y^2=16\) implicitly:

\(2x-6\left(x\frac{dy}{dx}+y\right)+50y\frac{dy}{dx}=0\).

At \((3,1)\), this becomes

\(6-6\left(3\frac{dy}{dx}+1\right)+50\frac{dy}{dx}=0\),

so \(32\frac{dy}{dx}=0\). Hence \(\frac{dy}{dx}=0\) at \((3,1)\).

Differentiate the first derivative equation again:

\(2-6\left(x\frac{d^2y}{dx^2}+2\frac{dy}{dx}\right)+50\left[\left(\frac{dy}{dx}\right)^2+y\frac{d^2y}{dx^2}\right]=0\).

At \((3,1)\), with \(\frac{dy}{dx}=0\), this gives

\(2-18\frac{d^2y}{dx^2}+50\frac{d^2y}{dx^2}=0\).

Thus \(2+32\frac{d^2y}{dx^2}=0\), so \(\frac{d^2y}{dx^2}=-\frac1{16}\).

Since the second derivative is negative, the turning point is a local maximum.