Hi Oleg- If the concept is something you're struggling with as a whole, I would recommend taking some time going through it in detail rather than trying to get it from one single problem or two. Please feel free to message me and we can take some time to do a quick online session to get you fully understanding!
With that said I hope the following solution is useful in helping you get the clarity you need!
So we want to find an equation of the tangent line @y=-1. If we want to find the equation of the line we will need 2 things: a slope and a point that the line goes through. To find the slope we will want to find the derivative. We have x's and y's so we will want to use implicit differentiation (we will find dy/dx). Since we are taking the derivative of y with respect to x, remember to tack on the dy/dx when differentiating y variables. If you wanted dx/dy that would just be the reciprocal of what we found.
11y2+yx3=3
22y dy/dx+3yx2+x3 dy/dx=0
Factoring the dy/dx out of the terms and moving what doesn't have it to the other side we get dy/dx[22y+x3]=-3yx2
Dividing we get dy/dx=-3yx2/[22y+x3]
Now, we know y=-1, but we also need an x value. If we go back to our original, we can plug in y=-1 and solve for x.
11y2+yx3=3
11(-1)2+(-1)x3=3
11-x3=3
Solving for x we get x3=8 and therefore, x=2.
So our point is (2,-1)
Plugging this into our dy/dx, we get dy/dx=-3yx2/[22y+x3]=-3(-1)(2)2/[22(-1)+(2)3]=-12/14=-6/7
With this as our slope and our point being (2,-1) we can use point-slope formula to find the equation of the line
The formula is y-y1=m(x-x1)
y-(-1)=-6/7(x-2)
y+1=(-6/7)x+12/7
y=(-6/7)x+5/7
Hope this helped!
