Answer:

Step-by-step explanation:
We have the equation:

And we want to find d²y/dx² at the point (-2, -1).
So, let's take the derivative of both sides with respect to x:
![\frac{d}{dx}[2y^2+2]=\frac{d}{dx}[x^2]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5B2y%5E2%2B2%5D%3D%5Cfrac%7Bd%7D%7Bdx%7D%5Bx%5E2%5D)
On the left, let's implicitly differentiate:
![4y\frac{dy}{dx}=\frac{d}{dx}[x^2]](https://tex.z-dn.net/?f=4y%5Cfrac%7Bdy%7D%7Bdx%7D%3D%5Cfrac%7Bd%7D%7Bdx%7D%5Bx%5E2%5D)
Differentiate normally on the left:

Solve for the first derivative. Divide both sides by 4y:

Now, let's take the derivative of both sides again:
![\frac{d}{dx}[\frac{dy}{dx}]=\frac{d}{dx}[\frac{x}{2y}]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5B%5Cfrac%7Bdy%7D%7Bdx%7D%5D%3D%5Cfrac%7Bd%7D%7Bdx%7D%5B%5Cfrac%7Bx%7D%7B2y%7D%5D)
We will need to use the quotient rule:
![\frac{d}{dx}[f/g]=\frac{f'g-fg'}{g^2}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5Bf%2Fg%5D%3D%5Cfrac%7Bf%27g-fg%27%7D%7Bg%5E2%7D)
So:
-x\frac{d}{dx}[(2y)]}{(2y)^2}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%3D%5Cfrac%7B%5Cfrac%7Bd%7D%7Bdx%7D%5B%28x%29%5D%282y%29-x%5Cfrac%7Bd%7D%7Bdx%7D%5B%282y%29%5D%7D%7B%282y%29%5E2%7D)
Differentiate:

Simplify:

Substitute x/2y for dy/dx. This yields:

Simplify:

Simplify. Multiply both the numerator and denominator by 2y. So:

Reduce. Therefore, our second derivative is:

We want to find the second derivative at the point (-2, -1).
So, let's substitute -2 for x and -1 for y. This yields:

Evaluate:

Multiply:

Subtract:

Reduce. So, our answer is:

And we're done!