y sin 3x , y³ cos x+5=0 find dy/dx

The product rule states $\frac{d}{dx}(uv) = u'v + uv'$.
Let $u = y^3$ and $v = \cos x$.
Then $u' = \frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx}$ (using the chain rule).
And $v' = \frac{d}{dx}(\cos x) = -\sin x$.
The derivative of the constant 5 is 0. The derivative of 0 is 0.

Substituting these into the equation:

$$(3y^2 \frac{dy}{dx})(\cos x) + (y^3)(-\sin x) + 0 = 0$$ $$3y^2 \cos x \frac{dy}{dx} - y^3 \sin x = 0$$

$$\frac{dy}{dx} = \frac{y \sin x}{3 \cos x}$$

This can also be written using the identity $\tan x = \frac{\sin x}{\cos x}$:

$$\frac{dy}{dx} = \frac{y}{3} \tan x$$