sorry i entered it wrong the first time
e^(x+y)+x-y^2=3
All we do for this is differentiate implicitly with respect to x, treating y as a separate variable:
We have:
$\displaystyle e^{x + y} + x - y^2 = 3$
Now we differentiate:
$\displaystyle \frac{d}{dx}[e^{x + y}] + 1 - \frac{d}{dx}[y^2] = 0$
$\displaystyle \frac{d}{dx}[e^x]\frac{d}{dx}[e^y] - 2y\frac{dy}{dx} = -1$
When we differentiate y, we differentiate it just like we would x, but we add a $\displaystyle \frac{dy}{dx}$ after it:
$\displaystyle e^x(e^y\frac{dy}{dx}) - 2y\frac{dy}{dx} = -1$
$\displaystyle \frac{dy}{dx}(e^{x + y} - 2y) = -1$
$\displaystyle \frac{dy}{dx} = -\frac{1}{e^{x + y} - 2y}$
And there you go.
That is not true. The product rule needs to be used if you were to go about it that way: $\displaystyle \frac{d}{dx} e^{x+y} = \frac{d}{dx} \left[e^{x}e^{y}\right] = \left(e^{x}\right)'e^y + e^{x}\left(e^y\right)' = \hdots$
However, it is sufficient to simply know that: $\displaystyle \frac{d}{dx}e^{u(x)} = e^{u(x)} \cdot u'(x)$. Here, we see that u = x + y, so:
$\displaystyle e^{x + y} + x - y^2 = 3$
$\displaystyle (x+y)'e^{x+y} + 1 - 2yy' = 0$
$\displaystyle (1 + y')e^{x+y} - 2yy' = -1$
etc. etc.