Finding the total differential. My understanding of this is that it is a rate of change and in my mind I think of it like a vector.. you go some units in the $x$ direction and some units in the $y$ direction giving you the total rate of change for a unit movement of $x$ and a unit movement of $y$. Is this correct ?
So if we start at some origin then the total differential is basically the distance we moved multiplied by the differential in that direction ?
When I solve this problem myself I get 
$$(2x + y)dx + (2y + x)dy$$ where wolfram says :
$$x\frac{dy}{dx} +2 y \frac{dy}{dx}+2 x+y$$
Are the two answers the same and could you please explain and could you please explain if my understanding of total differential is correct ?
Thank you,
 A: Please go through this playlist for a clearer understanding of differentiation.
You can learn about implicit derivatives like the one you've tried to do.

I'm not sure how you solved it but here's how I would it:

$$\frac{d}{dx} (x^2 + xy + y^2)\\
=\frac{d}{dx}(x^2) + \frac{d}{dx}(xy) + \frac{d}{dx}(y^2)\\
= 2x + x\frac{dy}{dx} + y\frac{dx}{dx} + 2y\frac{dy}{dx}\\
\equiv x\frac{dy}{dx} +2 y \frac{dy}{dx}+2 x+y \tag{W$\alpha$ is right!}$$
Here we've used two important concepts:


*

*$(uv)' = u'v + uv' \quad\dots$ (used in $\frac{d}{dx} xy$)

*$(f(g(x)))' = f'(g(x))\cdot g'(x) \quad\dots$ (used in $\frac{d}{dx} y^2$) 



Now there's a rough way of differentiating it without respect to any variable:
$$\Delta (x^2 + xy + y^2) \\
= 2x\Delta x + y\Delta x + x\Delta y + 2y\Delta y \\
= (2x + y)\Delta x + (2y + x)\Delta y \tag{You're right!}$$

So, basically you did do it right. Wolfram alpha (and I) did it with respect to $x$ and considered $y$ to be a function of $x$. 
You've basically differentiated with respect to a third variable. The variable itself is irrelevant in the way we've done it.

Mhh, maybe your understanding of differentiation is better than mine since you've got an answer with a self-made intuitive explanation :D
Good Job!
