If $y = \sqrt[3]{\tan x + y}$, then $\frac{dy}{dx} =$
- A$\frac{\sec^2 x}{3y^2 + 1}$
- B$\frac{\sec^2 x}{3y^2 - 1}$
- C$\frac{\tan x}{3y^2 - 1}$
- D$\frac{\sec^2 x}{3y - 1}$
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