Isn't partial differentiation implied by function context? This has been bothering me for some time, so I thought I'd finally ask it here. If we are given a function, say, $$f(x,y)=x^2+y^2,$$ and are asked to differentiate it w.r.t. $x$, i.e. $$\frac{\partial}{\partial x} f(x,y),$$ then surely this is the same as saying $$\frac{d}{dx} f(x,y),$$ since we will know by the context that obviously $$\frac{d}{dx} y^2 = 0,$$ and therefore $$\frac{\partial}{\partial x} f(x,y)=\frac{d}{dx} f(x,y)=2x.$$ Or am I missing something?
 A: You are indeed missing something:
$\frac{d}{dx}y^2 \ne0$
since $\frac{d}{dx}y^2 = 2y \frac{dy}{dx}$
Note: $\frac{d}{dx} = \frac{d}{dy} \frac{dy}{dx}$
$$\frac{d}{dx}y^2 = \frac{d}{dy} (y^2) \frac{dy}{dx}$$
$$=2y \frac{dy}{dx}$$
This is often referred to as implicit differentiation, and can prove helpful in finding $\frac{dy}{dx}$ in certain cases where direct differentiation is more difficult, eg: $x^2 - xy + y^3 = 8$.

$$x^2 - xy + y^3 = 8$$
$$\frac{d}{dx}(x^2) - \frac{d}{dx}(xy) + \frac{d}{dx}(y^3) = \frac{d}{dx}(8)$$
$$2x -x \cdot \frac{dy}{dx} - y + 3y^2 \cdot \frac{dy}{dx} = 0$$ Where we have used:  $$u=-x,v=y$$$$u'=-1,v'=\frac{dy}{dx}$$
$$\frac{dy}{dx}(3y^2 - x) = -2x + y$$
$$\frac{dy}{dx} = \frac{y -2x}{3y^2 - x}$$
by implicit differentiation.
Whereas by partial differentiation with respect to x, we would have obtained:
$$2x - y = 0$$
Which as you can see happens to be the top value of our implicit differential multiplied by $(-1)$, this is because implicit differentiation can be thought of as $-\frac{f_x}{f_y}$, being the partial derivative with respect to x divided by the partial with respect to y multiplied by $(-1)$.
