Prove that the second order implicit multivariable derivative equals $0$

Question

Let $ y $ be determinaded implicitly as a function of $ x $ by the equation $ x^2 + y^2 + 2axy = $ 0, with $ a > 1 $. Prove that $ \frac{d^2y}{dx^2} = 0 $.

My failed attempt

$$ \frac{dy}{dx} = - \dfrac{\frac{\partial z}{\partial x}}{\frac{\partial z}{\partial y}} $$

$$ \frac{dy}{dx} = -\frac{z_x}{z_y} $$

$$ \frac{d^2y}{dx^2} = \frac{d}{dx} \Biggl[ \frac{-z_x}{z_y} \Biggr] = \dfrac{\frac{d}{dx}[-z_x](z_y) + (z_x)\frac{d}{dx}[z_y]}{(z_y)^2} $$

Now, $ z_x = 2x + 2ay$ and $ z_y = 2y + 2ax $. Therefore,

$$ \frac{d}{dx}[-z_x] = \frac{d}{dx}[-2x -2ay] = -2 $$

$$ \frac{d}{dx}[z_y] = \frac{d}{dx}[2ax + 2y] = 2a $$

It follows that

$$ \frac{dy^2}{dx^2} = \frac{4a^2y - 4y}{(2y + 2ax)^2} $$

And that numerator just isn't equal to $0$ since $a > 1$.

Any help is appreciated.

Edit: After using Dr. Sonnhard's tip, that

$$ y''=-\frac{(ay'+1)(y+ax)-(ay+x)(y'+a)}{(y+ax)^2} $$

You can expand everything, and simplify, and you still don't get a $0$ because the terms in the numerator don't cancel each other out. I'm thinking that there is a problem with the question.

Answer 1

I think the question should have been to prove $$\frac{d^2y}{dx^2} > 0 $$ So that if we take 4y as common from your last equation, $$ 4a^2y - 4y = 4y(a^2 - 1)$$ And as, $$ a>1, a^2>1 , a^2 - 1 > 0$$ If y>0, then $$\frac{d^2y}{dx^2} > 0$$ Thank you.

Answer 2

Hint: For the first derivative i got: $$y'=-\frac{ay+x}{y+ax}$$ and differentiating again $$y''=-\frac{(ay'+1)(y+ax)-(ay+x)(y'+a)}{(y+ax)^2}$$ and for $$y'$$ you have to substitute the first expression.

Answer 3

The given equation can be rewritten as $$\bigl(y+ax+\sqrt{a^2-1}\ x\bigr)\cdot\bigl(y+ax-\sqrt{a^2-1}\ x\bigr)=0\ .\tag{1}$$ Its solution set is the union of the two lines $$y=\bigl(-a\pm\sqrt{a^2-1}\bigr)\>x\ .$$ In the neighborhood of any point $P\ne(0,0)$ on such a line the local function $x\mapsto y(x)$ defined by $(1)$ therefore has $${d^2y\over dx^2}\equiv0\ .$$