Given:
The equation is:

To find:
The equation of the tangent on the given equation at point (-3,2).
Solution:
We have,

Differentiate with respect to x.
![4(1)+2[x(2yy')+y^2(1)]+0=3y^2y'](https://img.qammunity.org/2022/formulas/mathematics/high-school/r4gafp2dnnee4i962z25d8s33zr9zcpou2.png)



Isolate y'.

Now, we need to find the value of the derivative at point (-3,2).




It means the slope of the tangent line is
.
The equation of tangent line that passes through the point (-3,2) with slope
is:





Therefore, the equation of tangent line is
.