Answer: At \(A(2,0)\),

\(\frac{\mathrm dy}{\mathrm dx}=2,\qquad \frac{\mathrm d^2y}{\mathrm dx^2}=-9.\)

The curve is

\(x^2-xy-2y^2=4.\)

Differentiate implicitly:

\(2x-(xy'+y)-4yy'=0.\)

Thus

\(2x-y=(x+4y)y'.\)

So

\(y'=\frac{2x-y}{x+4y}.\)

At \(A(2,0)\),

\(y'=\frac{4}{2}=2.\)

Now differentiate

\(2x-y=(x+4y)y'\)

again:

\(2-y'=(1+4y')y'+(x+4y)y''.\)

At \(x=2\), \(y=0\), \(y'=2\),

\(2-2=(1+8)(2)+2y''.\)

So

\(0=18+2y'',\)

and therefore

\(y''=-9.\)