1. d^2y/dx^2

find $d^2y/dx^2$ by implicit differentiation

$3x^2-4y^2=7$

$6x-8yy'=0$

$y'=\frac{6x}{8y}=-\frac{3x}{4y}$

$\frac{d^2y}{dx^2} = \frac{3x(4y')-4y(3)}{16y^2}$

i don't see this moving towards $-\frac{21}{16y^3}$which is the answer

2. In your formula for $\frac{d^2y}{dx^2}$, you have a y', which you can replace by what you found just above ! : $y'=-\frac{3x}{4y}$
And then finish it off with the original equation : $3x^2-4y^2=7 \Rightarrow x^2=\frac{7+4y^2}{3}$

Plus, you applied the quotient rule incorrectly, the derivative of $\frac uv$ is $\frac{u'v-uv'}{v^2}$, but you did $uv'-u'v$

3. Plus, you applied the quotient rule incorrectly, the derivative of $\frac {u}{v}$ is $\frac{u'v-uv'}{v^2}$, but you did $uv'-u'v$
then
$\left(-\frac{3x}{4y}\right)' = -\frac{(3)4y-3x(4y')}{16y^2}$

4. Originally Posted by bigwave
then
$\left(-\frac{3x}{4y}\right)' = -\frac{(3)4y-3x(4y')}{16y^2}$
So sorry, there was indeed the minus sign...
Did you manage to go through all the calculations to get the desired answer ?

5. We have $3x^2-4y^2=7$ and $y'=-\frac{3x}{4y}$

$\frac{d^2y}{dx^2}=\frac{3x(4y')-4y(3)}{16y^2}=\frac{12(xy'-y)}{16y^2}=\frac 34\cdot\frac{xy'-y}{y^2}$

But $y'=-\frac{3x}{4y}\Rightarrow xy'=-\frac{3x^2}{4y}\Rightarrow xy'-y=-\frac{3x^2}{4y}-y=-\frac{3x^2-4y^2}{4y}$

But the very first equation tells us that $3x^2-4y^2=7$

Hence $xy'-y=-\frac{7}{4y}$

Therefore, $\frac{d^2y}{dx^2}=\frac 34\cdot\left(-\frac{7}{4y^3}\right)=-\frac{21}{16y^3}$