This is a matter of understanding what you're dealing with.
You're asked to differentiate $x^2+y^2=1$. An equation isn't a differentiable function, therefore the equation can't be differentiated.
Now comes the 'translating the problem part'.
The equation $x^2+y^2=1$ 'defines a function', more precisely, there exists a function $g\colon U\to V$ such that $x^2+(g(x))^2=1$, for some sets $U$ and $V$. (A lot can be said about $g, U$ and $V$). Let's assume for the time being that $g$ is differentiable. Now what the problems is actually asking you to do is to differentiate both sides of $x^2+(g(x))^2=1$, yielding $2x+2g(x)g'(x)=0$.
All this is simply the Implicit Function Theorem. The details can be checked on the link.
In two dimensions the theorem goes as follows:
Let $D\subseteq \Bbb R^2$ be an open set and let $f\colon D \to \Bbb R$ be a class $C^1$ function. Given $a\in \Bbb R$, suppose there exists $(x_0, y_0)\in D$ such that $f(x_0, y_0)=c$ and $f_y(x_0, y_0)\neq 0$. Then there are open intervals $U$ and $V$ with the property that there exists a class $C^1$ function $g\colon U\to V$ such that $\forall x\in U\left(f(x,g(x))=c\right)$. Furthermore defining $h\colon U\to \Bbb R, x\mapsto f(x,g(x))$, the chain rule yields $\forall x\in U(h'(x)=f_x(x,g(x))+f_y(x,g(x))g'(x)=0)$.