What makes the second question seem intimidating is that we are used to being asked to find specific expressions for the various higher derivatives found by implicit differentiation, but that isn't called for here.
Sammy Black shows the start of the "chain" of implicit differentiations that we need to make on the equation of the curve. There are a couple of things we can do to make the resolution of the problem clearer. One is that we do not need to do even so much as to collect the terms in the derivative equations, because of the second item. Since we are asked about the behavior of the higher derivatives of the implicit function $ \ y \ = \ \varphi ( x ) \ $ in the neighborhood of the origin , the amount of effort we need to make is reduced immensely.
The first round of differentiation (as already shown) produces
$$ \frac{d}{dx} \ [ \ x \ + \ y \ + \ x^5 \ - \ y^5 \ ] \ = \ \frac{d}{dx} [ \ 0 \ ] \ \ \Rightarrow \ \ 1 \ + \ y \ ' \ + \ 5x^4 \ - \ 5y^4 \ y \ ' \ \ = \ 0 $$
$$ \Rightarrow \ \ 1 \ + \ 5x^4 \ + \ ( \ 1 \ - \ 5y^4 \ ) \ y \ ' \ \ = \ 0 \ \ . $$
We wish to evaluate the left-hand side at $ \ (x,y) \ = \ (0,0) \ $ , which gives us
$$ \Rightarrow \ \ 1 \ + \ 5 \cdot 0 \ + \ ( \ 1 \ - \ 5 \cdot 0 \ ) \ y \ ' \ \ = \ 0 \ \ \Rightarrow \ \ y \ ' \ = \ \varphi \ ' (0) \ = \ -1 \ \ . $$
We next implicitly differentiate our first-derivative equation repeatedly as it stands, and perform a similar evaluation each time:
$$ \frac{d}{dx} \ [ \ 1 \ + \ 5x^4 \ + \ ( \ 1 \ - \ 5y^4 \ ) \ y \ ' \ ] \ = \ \frac{d}{dx} [ \ 0 \ ] $$
$$ \Rightarrow \ \ 20x^3 \ + \ ( \ -20y^3 \cdot y \ ' \ ) \cdot y \ ' \ + \ ( \ 1 \ - \ 5y^4 \ ) \ y \ '' \ \ = \ 0 $$
$$ \Rightarrow \ \ \text{at (0,0)} : \ \ 20 \cdot 0 \ + \ ( \ 20 \cdot 0 \cdot y \ ' \ ) \cdot y \ ' \ + \ ( \ 1 \ - \ 5 \cdot 0 \ ) \ y \ '' \ \ = \ 0 $$
$$ \Rightarrow \ \ y \ '' \ = \ \varphi \ '' (0) \ = \ 0 \ \ ; $$
$$ \frac{d}{dx} \ [ \ 20x^3 \ + \ ( \ -20y^3 \ ) \cdot ( \ y \ ' \ )^2 \ + \ ( \ 1 \ - \ 5y^4 \ ) \ y \ '' \ ] \ = \ \frac{d}{dx} [ \ 0 \ ] $$
$$ \Rightarrow \ \ 60x^2 \ + \ ( \ -60y^2 \cdot y \ ' \ ) \cdot ( \ y \ ' \ )^2 \ + \ ( \ -20y^3 \ ) \cdot ( \ 2 y \ ' \ y \ '' \ ) \ + \ ( \ - 20y^3 \cdot y \ ' \ ) \ y \ '' \ + \ ( \ 1 \ - \ 5y^4 \ ) \ y \ ''' \ \ = \ 0 $$
$$ \Rightarrow \ \ \text{at (0,0)} : \ \ 60 \cdot 0 \ - \ 60 \cdot 0 \cdot 0 \ - \ 20 \cdot 0 \cdot 0 \ - \ 20 \cdot 0 \cdot 0 \ + \ ( \ 1 \ - \ 5 \cdot 0 \ ) \ y \ ''' \ \ = \ 0 $$
$$ \Rightarrow \ \ y \ ''' \ = \ \varphi \ ''' (0) \ = \ 0 \ \ . $$
At this point, we notice something that will relieve the tedium of computing all those products of higher derivatives: except for the first and last terms on the left-hand side, all the others involve only $ \ y \ $ and its derivatives, all of which so far have proven to equal zero. So for the successive higher derivatives of the left-hand side of the equation, we can just focus on the terms that behave otherwise:
fourth derivative --
$$ 120x \ + \ \ldots \ + \ ( \ 1 \ - \ 5y^4 \ ) \ y^{(4)} \ \ = \ 0 \ \ \Rightarrow \ \ y^{(4)} \ = \ \varphi^{(4)} (0) \ = \ 0 \ \ $$
[still nothing interesting, but we can see what's coming...]
fifth derivative --
$$ 120 \ + \ \ldots \ + \ ( \ 1 \ - \ 5y^4 \ ) \ y^{(5)} \ \ = \ 0 \ \ \Rightarrow \ \ y^{(5)} \ = \ \varphi^{(5)} (0) \ = \ -120 \ \ \text{(!)} $$
We at last reach a result that perhaps ends our wondering why the implicit function isn't simply $ \ y \ = \ -x \ $ . What we learn from this is that the curve described by our equation is extremely flat in the neighborhood of the origin:
in this enlargement, the red line ( y = -x ) indicates where our curve begins to deviate significantly from a straight line
As for still higher derivatives, it is now apparent that, with the term involving $ \ x \ $ "differentiated away", the only non-zero derivatives of $ \ y \ $ will only be found in terms which are products with other factors which equal zero. We may immediately conclude from this that $ \ \varphi^{(k)} (0) \ = \ 0 \ $ , for $ \ k \ \ge \ 6 \ . $
[Note that this generalizes easily for equations $ \ x \ + \ y \ + \ x^n \ - \ y^n \ = \ 0 \ $ , for odd $ \ n \ $ . When $ \ n \ $ is even, the equation describes the union of a curve which does not pass through the origin and the "degenerate" line $ \ y \ = \ -x \ $ , for which all derivatives beyond $ \ y \ ' \ $ are zero at the origin.]
