Implicit Differentiation Doubt One of my classmates said that for $x^2+y^2=1$, to find $\frac{dy}{dx}$, the following method can be used: 
First rearrange the equation, $$x^2+y^2-1=0$$. Then assume $$w=x^2+y^2-1$$ $$\frac{dy}{dx}=\frac{dw}{dx} \div \frac{dw}{dy}$$
Also, when finding $\frac{dw}{dx}$, $y$ is considered a constant. And similarly, when finding $\frac{dw}{dy}$, $x$ is considered a constant. 
Let me proceed.
$\frac{dw}{dx}=2x$ regarding $y$ as a constant.
$\frac{dw}{dy}=2y$ regarding $x$ as a constant.
$\frac{dy}{dx}=2x/2y$ which is the negative of the correct answer.
It seems like it applies for all the equations.
I think his methods is completely unreasonable but by using his method the result is always the negative of the correct result. Why????
 A: What he is showing you is the result of partial differentiation.  In partial differentiation, for a function $$u = f(x, y),$$ then $$du = \partial_x u + \partial_y u$$ where $\partial_x u$ is the partial differential of $u$ with respect to $x$ (i.e., differentiating with all variables but $x$ held constant).
Since, in your case, $u$ (and therefore $du$) is zero, we can then say:
$$ 0 = \partial_x u + \partial_y u $$
Now, if we solve for one of them, we get:
$$ \partial_x u = -\partial_y u $$
Or, in other words,
$$ \frac{\partial_x u}{\partial_y u} = -1 $$
That's for differentials.  The partial derivative of $u$ with respect to $x$ is actually $\frac{\partial_x u}{dx}$.  Now, let's multiply both sides by $\frac{dy}{dx}$:
$$\frac{\partial_x u}{\partial_y u} \frac{dy}{dx} = -\frac{dy}{dx} $$
We can rearrange the left-hand side so that we get partial derivatives:
$$\frac{\frac{\partial_x u}{dx}}{\frac{\partial_y u}{dy}} = -\frac{dy}{dx} $$
And that is the procedure that your friend gave you.  However, for a general technique for implicit differentiation, the method that @MrYouMath above gives is generally more straightforward.
A: You do not have to rearrange the equation just apply the total differential to the equation
$$x^2+y^2=1 \implies d(x^2)+d(y^2)=d(1) \implies 2xdx+2ydy = 0.$$
Now, solve for $dy/dx$.
A: Similar to the other answers, I would differentiate with respect to $x$ and treat $y=y(x)$ by computing
\begin{align}
\frac{d}{dx}\left(x^2+y(x)^2\right) &= \frac{d}{dx}(1)\\
\implies 2x+2y\frac{dy}{dx} &= 0\\
\implies \frac{dy}{dx} &= -\frac{x}{y} \qquad \blacksquare
\end{align}
A: It is quite reasonable. Note:
$$x^2+y^2=1 \Rightarrow w(x,y)=x^2+y^2-1=0.$$
The total differential of the function $w(x,y)$ is:
$$w_xdx+w_ydy=0 \Rightarrow \frac{dy}{dx}=-\frac{w_x}{w_y}=-\frac{\frac{dw}{dx}}{\frac{dw}{dy}}.$$
So your classmate is only missing minus.
