More help with implicit differentiation Given $(5x+5y~)^3= 125x^3+125y^3$, find the derivative.
Using the chain rule and power rule, I came up with
$3(5x+5y)^2 \cdot (\frac{d}{dx}5x+\frac{dy}{dx}5y)= 3 \cdot 125x^2 +3 \cdot 125y^2$
Now, the derivative of $5x~$ is 5, but what about the derivative of $5y~$?  
I know that $\frac{dy}{dx}5y~$ turns to $5(\frac{dy}{dx}(y~))$
What happens after that?  When I plugged the formula into Wolfram Alpha to double check my steps, it says that $\frac{dy}{dx}(y~)=0$  What is the reasoning behind that?

This seems settled; I'll address how a problem like this would be handled in Mathematica (unfortunately the functionality doesn't work in Wolfram Alpha).
You'll first want to express your given in the form $f(x,y)=0$, like so:
expr = (5x + 5y)^3 - 125x^3 - 125y^3;

The key is to remember that Mathematica supports two sorts of derivatives: the partial derivative D[] and the total derivative Dt[]. As your y depends on x, Dt[] is what's appropriate here (D[] treats y as a constant, so D[y, x] gives 0).
Dt[expr, x]
-375 x^2 - 375 y^2 Dt[y, x] + 3 (5 x + 5 y)^2 (5 + 5 Dt[y, x])

From there, you can use Solve[] to solve for Dt[y, x] like so:
Dt[y, x] /. Solve[Dt[expr, x], Dt[y, x]]

