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?
 A: Others have pointed out that
$\frac{d}{dx}(5x+5y)^3 = 3 (5x + 5y)^2 \frac{d}{dx}(5x + 5y) = 3 (5x + 5y)^2 (5 + 5 \frac{dy}{dx})$
because $\frac{d}{dx}(5x + 5y) = \frac{d}{dx}(5x) + \frac{d}{dx}(5y) = 5 + 5 \frac{dy}{dx}$.  In your calculation when you write $\frac{dy}{dx} y$, it means $y$ multiplied with $\frac{dy}{dx}$.  When you write $\frac{d}{dx}(y)$, it means $\frac{dy}{dx}$.  You can think of $\frac{d}{dx}$ as a function that requires another function as input.  Just like a function $f$ requires a number as input so that's why $f(x)$ doesn't mean $x$ multiplied with $f$; it means $x$ is the input to $f$.
On the right hand side, you forgot to apply the Chain Rule:
$\frac{d}{dx}(125x^3 + 125y^3) = \frac{d}{dx}(125x^3) + \frac{d}{dx}(125y^3) = 375x^2 + 375y^2 \frac{dy}{dx}$.
Finally, after this you can solve for $\frac{dy}{dx}$ as usual by isolating $\frac{dy}{dx}$ to one side.
A: You are applying $\frac{d}{dx}$ to both sides.  On the left side, you should have $\frac{d}{dx}5y$ instead of $\frac{dy}{dx}5y$, which is just $5\frac{dy}{dx}$.  And when you take $\frac{d}{dx}125y^3$ on the right, you need to use the chain rule to involve $\frac{dy}{dx}$ as well.
A: 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]]

