Past exam of the mathematical methods II course of the University of Oxford / 2013 resit / 11 / c / ii

On $y=x$, the restricted function is $g(x)=2x^4-16x^2$. Its derivative is $g^{\prime }(x)=8x(x^2-4)$, so the turning points are $x=-2,0,2$. The restricted second derivative is $g^{\prime \prime }=24x^2-32$, negative at zero and positive at $\pm 2$. ThusThe corresponding points in the plane are $(0,0)$ and $(\pm 2,\pm 2)$. Along the diagonal the derivative is $f_x+f_y$, not a single arbitrary partial derivative; here $f_x=f_y$, so their zeros and sign changes agree with the restricted-curve calculation.