Differentiability question wouldn't $f(0,0)$ be undefined because Hi i was given a set of solutions i couldn't and got stuck at this part. Please My question is not the given question but rather the given solution.
I know that differentiability is to show that 
$$\lim_{h \to 0 } \frac{f(a+h,b+h)+f(a,b)}{h} $$
But the question is....

And then i'm told.... 

Which is to what i figured
But then there is this part...


wouldn't $f(0,0)$ be inconsistent? Also why "$0+t$" alternate between the two is it because all of y is zero along the path of x and vice versa for y? Why are we using the differentiability of each derivative?

 A: The key error you've made is the definition of differentiability.
A function of two real variable $f(x,y)$ is differentiable at $(a,b)$ if there are real numbers $u,v$, such that:
$$\lim_{(x,y)\to (a,b)} \frac{f(x,y)-f(a,b)-u(x-a)-v(y-b)}{\sqrt{(x-a)^2+(y-b)^2}} = 0$$ 
This can be read as saying, roughly:

$f(x,y)$ can be approximated well near $(a,b)$ as $$f(x,y)\approx f(a,b)+u(x-a)+v(y-b)$$

The denominator $\sqrt{(x-a)^2+(y-b)^2}$ indicates how good this approximation must be when we are near $(a,b)$. In particular, if $f$ is differentiable, then $u,v$ are distinct - there is at most one pair $(u,v)$ which matches this condition.
In all cases, if there is a pair $u,v$ then $$u=\frac{\partial f}{\partial x}(a,b), v=\frac{\partial f}{\partial x}(a,b)$$
But the point of this exercise/discussion is to show that $\frac{\partial f}{\partial x}(a,b)$ and $\frac{\partial f}{\partial y}(a,b)$ can exist and still not have $f$ differentiable at $(0,0)$.
In this particular case, $(a,b)=(0,0)$ and $f(x,y)=\frac{x^3-y^3}{x^2+y^2}$ and you can show that if $(u,v)$ exists, they must be the partial derivatives, $(u,v)=(1,-1)$ This yields, when $(x,y)\neq 0$:
$$\frac{f(x,y)-f(0,0)-ux-vy}{\sqrt{x^2+y^2}} = \frac{xy(y-x)}{(x^2+y^2)^{3/2}}$$
If $y=2x$ then we get:
$$\frac{xy(y-x)}{(x^2+y^2)^{3/2}} = \frac{2x^3}{|x|^3\cdot 5^{3/2}}=\pm \frac{2}{5^{3/2}}$$
So the left side does not converge to $0$ as $(x,y)\to 0$.
A: What is the definition of partial derivative of a map? Recall that a partial derivative of a map is a special case of the directional derivative of the map in the direction of a usual standard unit vector; this is why you see "alternation". 
If $f$ is differentiable at some $a \in \mathbb{R}^{2}$, i.e. if the derivative $df^{a}: \mathbb{R}^{2} \to \mathbb{R}$ of $f$ at $a$ exists, then we have
$$
f'(a; y) = df^{a}(y)
$$
for all $y \in \mathbb{R}^{2}$.
Using this theorem, let us see if $f$ is differentiable at $(0,0)$. If $y = (y_{1},y_{2}) \in \mathbb{R}^{2}$ such that $y_{1} \neq y_{2}$, then 
$$
h^{-1}\big[f(hy_{1},hy_{2}) - f(0,0) \big]= h^{-1}\frac{y_{1}^{3} - y_{2}^{3}}{y_{1}^{3} + y_{2}^{3}}
$$
for all $h \neq 0$, so $f'((0,0);y)$ does not exist, and hence $f$ is not differentiable at $(0,0)$.
