Use implicit differentiation to find $\frac{dc}{dv}$ when $v=\sqrt{c^2+v^2}$
I know I can solve this using normal implicit methods however I was wondering, why can I not square this so it becomes $v^2=c^2+v^2$ and then simplify so $ c^2=0 \implies c=0$ then the derivative of this is just $c'=0$, what is flawed?
There is no flaw in your method. Even if you use implicit differentiation, you will get : $$ dv=\frac{c dc}{\sqrt{c^2+v^2}}+\frac{v dv}{\sqrt{c^2+v^2}}$$ now use $\sqrt{c^2+v^2}=v$ to get $$dv=\frac{c}{v}dc+dv \implies dc=0$$
Which is the same as you got squaring both the sides.
Notice that $c(v)=0$, because $v^2=c^2+v^2$ directly implies $c=0$.
Thus, $\displaystyle \frac{dc}{dv}=0$. Your reasoning is fine.
