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Please help. I'm so confused
$\displaystyle f(x) = 3x^2 - 4xy + 3y^3 = -1.3 e + 3$
$\displaystyle f'(x) = 6x - 4y - 4x \frac{dy}{dx} + 9 y^2 \frac{dy}{dy} = -1.3e$
$\displaystyle \implies 6x - 4y +1.3e = 4x \frac{dy}{dx} - 9y^2 \frac{dy}{dy}$
$\displaystyle 6x - 4y +1.3e = \frac{dy}{dx} (4x - 9y^2)$
$\displaystyle \frac{dy}{dx} = \frac{6x - 4y + 1.3e}{4x - 9y^2}$
Gradient found when $\displaystyle (5 , -8)$ substituted into $\displaystyle \frac{dy}{dx}$.
$\displaystyle Gradient = -0.118$ to $\displaystyle 3sf$
Equation of tangent:
$\displaystyle y - (-8) = -0.118 (x - 5)$
$\displaystyle y = -0.118x -7.41$