If y = x tan y, then dy/dx =

The correct answer is $\frac{y}{x-x^2-y^2}$.

Let us find the derivative $\frac{dy}{dx}$ of the implicit equation given by:
$$y=x\tan \left(y\right)$$

Differentiating both sides with respect to $x$ using the product rule on the right-hand side, we get:
$$\frac{dy}{dx}=\frac{d}{dx}\left[x\right]\cdot \tan \left(y\right)+x\cdot \frac{d}{dx}[\tan (y\left)\right]$$

Applying the chain rule for the derivative of $\tan \left(y\right)$:
$$\frac{dy}{dx}=\tan \left(y\right)+x{\sec }^2\left(y\right)\frac{dy}{dx}$$

Now, collect all terms containing $\frac{dy}{dx}$ on the left side of the equation:
$$\frac{dy}{dx}-x{\sec }^2\left(y\right)\frac{dy}{dx}=\tan \left(y\right)$$

Factor out $\frac{dy}{dx}$:
$$\frac{dy}{dx}(1-x{\sec }^2(y\left)\right)=\tan \left(y\right)$$

Solve for $\frac{dy}{dx}$:
$$\frac{dy}{dx}=\frac{\tan \left(y\right)}{1-x{\sec }^2\left(y\right)}$$

From the original equation $y=x\tan \left(y\right)$, we can write:
$$\tan \left(y\right)=\frac{y}{x}$$

Using the trigonometric identity ${\sec }^2\left(y\right)=1+{\tan }^2\left(y\right)$, we substitute this expression in:
$${\sec }^2\left(y\right)=1+{\left(\frac{y}{x}\right)}^2=1+\frac{y^2}{x^2}=\frac{x^2+y^2}{x^2}$$

Now substitute these values back into the expression for $\frac{dy}{dx}$:
$$\frac{dy}{dx}=\frac{\frac{y}{x}}{1-x\left(\frac{x^2+y^2}{x^2}\right)}$$

Simplify the denominator:
$$1-\frac{x^2+y^2}{x}=\frac{x-(x^2+y^2)}{x}=\frac{x-x^2-y^2}{x}$$

Now substitute this back to obtain the final simplified derivative:
$$\frac{dy}{dx}=\frac{\frac{y}{x}}{\frac{x-x^2-y^2}{x}}=\frac{y}{x-x^2-y^2}$$