a) State $2y\frac{{dy}}{{dx}}$ as derivative of ${y^2}$, or equivalent
Equate derivative of LHS to zero and solve for $\frac{{dy}}{{dx}}$
Obtain given answer correctly
b) Equate gradient expression to $ - 1$ and rearrange
Obtain $y = 2x$
Substitute into original equation to obtain an equation in ${x^2}$ (or ${y^2}$)
Obtain $2{x^2} - 3x - 2 = 0$ (or ${y^2} - 3y - 4 = 0$)
Correct method to solve their quadratic equation
State answers $\left( { - {\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}},{\text{ }} - 1} \right)$ and $\left( {2,{\text{ }}4} \right)$