What is $\frac{d^2y}{dx^2}$ of $(y^5) + (x^5) = 8$ when you solve using implicit differentiation? 
Find the Second derivative with respect to $x$ of :
$$ y^5 + x^5 = 8$$

When I solved it I got $\frac yx$ as the second derivative but I don’t think that I am right. Could someone explain it to me.
 A: we get for $$y(x)\neq 0$$ $$y'(x)=-\frac{x^4}{y^4}$$ so
$$y''(x)=-\frac{4x^3y^4-x^44y^3y'}{y^8}$$
plugging $$y'=-\frac{x^4}{y^4}$$ we get
$$y''=-\frac{4(x^3y^5+x^8)}{y^{9}}$$
A: Starting with
$y^5 + x^5 = 8, \tag 1$
we differentiate once to obtain
$5y^4 y' + 5x^4 = 0, \tag 2$
from which we may isolate $y'$:
$y' = -\dfrac{x^4}{y^4}; \tag 3$
differentiating this expression using the quotient rule yields
$y'' = -\dfrac{4x^3y^4 - 4x^4y^3y'}{y^8} = -\dfrac{4x^3y - 4x^4y'}{y^5}; \tag 4$
substitute (3) into (4):
$y'' = -\dfrac{4x^3y - 4x^4(-x^4/y^4)}{y^5} = -\dfrac{4x^3y^5 + 4x^8}{y^9}; \tag 5$
we can now in fact eliminate $y$ from the numerator using (1) in the form
$y^5 = 8 - x^5, \tag 6$
that is
$y'' = -\dfrac{4x^3y^5 + 4x^8}{y^9}$
$= -\dfrac{4x^3(8 - x^5) + 4x^8}{y^9} = -\dfrac{32x^3 - 4x^8 + 4x^8}{y^9} = -\dfrac{32x^3}{y^9}; \tag 7$
it is possible to use (6) once again to reduce the power of $y$ in the denominator, viz.
$y'' = -\dfrac{32x^3}{y^9} = -\dfrac{32x^3}{(8 - x^5)y^4}; \tag 8$
an expression which may have some utility, though I am somewhat skeptical as to its genuine advantage over, say, (7).
A: Since I'm not completely satisfied with the other answers, rewrite the equation as
$$y(x)^5=8-x^5$$
and now partially differentiate with respect to 
$x$ on both sides (using the chain rule):
$$5y^4(x)y'(x)=-5x^4$$
and dividing by $y^4(x)$ you obtain
$$y'(x)=-\frac{x^4}{y^4(x)}$$
and so on and so forth
