For the curve x+y=3xy which of the following is equal to (dy)/(dx) (3y)/(1-3x) (3y-1)/(1-3x) (3y+1)/(3x+1) (1)/(3x+1)

To find the derivative of \( y \) with respect to \( x \) for the curve \( x + y = 3xy \), we need to implicitly differentiate both sides of the equation with respect to \( x \).<br /><br />Starting with the given equation:<br />\[ x + y = 3xy \]<br /><br />Differentiate both sides with respect to \( x \):<br />\[ \frac{d}{dx}(x + y) = \frac{d}{dx}(3xy) \]<br /><br />Using the chain rule on the left side:<br />\[ 1 + \frac{dy}{dx} = 3 \left( x \frac{dy}{dx} + y \right) \]<br /><br />Simplify the right side:<br />\[ 1 + \frac{dy}{dx} = 3x \frac{dy}{dx} + 3y \]<br /><br />Rearrange the equation to isolate \(\frac{dy}{dx}\):<br />\[ 1 + \frac{dy}{dx} = 3x \frac{dy}{dx} + 3y \]<br />\[ 1 = 3x \frac{dy}{dx} + 3y - \frac{dy}{dx} \]<br />\[ 1 = (3x - 1) \frac{dy}{dx} + 3y \]<br /><br />Isolate \(\frac{dy}{dx}\):<br />\[ 1 - 3y = (3x - 1) \frac{dy}{dx} \]<br />\[ \frac{dy}{dx} = \frac{1 - 3y}{3x - 1} \]<br /><br />Thus, the correct answer is:<br />\[ \frac{3y-1}{1-3x} \]<br /><br />So, the correct choice is:<br />\[ \frac{3y-1}{1-3x} \]