The right hand side is a product of (cos x)3 and (tan x).

Now (cos x)3 is a power of a function and so we use Differentiating Powers of a Function:

`d/(dx)u^3=3u^2(du)/(dx)`

With u = cos x, we have:

`d/(dx)(cos\ x)^3=3(cos\ x)^2(-sin\ x)`

Now, from our rules above, we have:

`d/(dx)tan\ x=sec^2x`

Using the Product Rule and Properties of tan x, we have:

`(dy)/(dx)`

`=[cos^3x\ sec^2x]` `+tan\ x[3(cos\ x)^2(-sin\ x)]`

`=(cos^3x)/(cos^2x)` `+(sin\ x)/(cos\ x)[3(cos\ x)^2(-sin\ x)]`

`=cos\ x-3\ sin^2x\ cos\ x`

We need to determine if this expression creates a true statement when we substitute it into the LHS of the equation given in the question.

` "LHS"`

`=cos\ x(dy)/(dx)` `+3y\ sin\ x-cos^2x`

`=cos\ x(cos\ x-3\ sin^2x\ cos\ x)` `+3(cos^3x\ tan\ x)sin\ x-cos^2x`

`=cos^2x` `-3\ sin^2x\ cos^2x` `+3\ sin^2x\ cos^2x` `-cos^2x`

`=0`

` ="RHS"`

We have shown that it is true.