Answer: The stationary points are \((2,-2)\) and \((-2,2)\). The point \((2,-2)\) is a local maximum, and \((-2,2)\) is a local minimum.

Differentiate the curve equation implicitly with respect to \(x\):

\(x^{2}+2xy-4y^{2}+20=0\)

gives

\(2x+2\left(x\frac{dy}{dx}+y\right)-8y\frac{dy}{dx}=0\).

Collect the terms containing \(\frac{dy}{dx}\):

\(2x+2y+\frac{dy}{dx}(2x-8y)=0\).

So

\(\frac{dy}{dx}(x-4y)=-(x+y)\).

If the tangent is parallel to the \(x\)-axis, then its gradient is zero, so \(\frac{dy}{dx}=0\). Substituting into the differentiated equation gives

\(x+y=0\), hence \(y=-x\).

Now substitute \(y=-x\) into the equation of the curve:

\(x^{2}+2x(-x)-4(-x)^{2}+20=0\)

\(x^{2}-2x^{2}-4x^{2}+20=0\)

\(-5x^{2}+20=0\)

\(x^{2}=4\), so \(x=\pm 2\).

Since \(y=-x\), the corresponding points are

\((2,-2)\) and \((-2,2)\).

To determine the nature of these stationary points, differentiate \(\frac{dy}{dx}(x-4y)=-(x+y)\) again with respect to \(x\):

\(\frac{d^{2}y}{dx^{2}}(x-4y)+\frac{dy}{dx}\left(1-4\frac{dy}{dx}\right)=-1-\frac{dy}{dx}\).

At a stationary point, \(\frac{dy}{dx}=0\), so this simplifies to

\(\frac{d^{2}y}{dx^{2}}(x-4y)=-1\).

For \((2,-2)\), we have \(x-4y=2-4(-2)=10\), so

\(\frac{d^{2}y}{dx^{2}}=-\frac{1}{10}\), which is negative, hence \((2,-2)\) is a local maximum.

For \((-2,2)\), we have \(x-4y=-2-8=-10\), so

\(\frac{d^{2}y}{dx^{2}}=\frac{1}{10}\), which is positive, hence \((-2,2)\) is a local minimum.