Find equation of Tangent line at $(4, 1)$ on $5y^3 + x^2 = y + 5x$ Can someone help me find equation of tangent line at $(4, 1)$ on
 $5y^3 + x^2 = y + 5x$
$Y=f(x)$
I dont know how to isolate the $Y$

knowing
$$
y-y_0 = y'(x-x_0)
$$
$$
\frac{d}{dx}[3y^3+x^2] = \frac{d}{dx}[y+5x]
$$
$$
15y^2 y' + 2x = y' + 5
$$
$$
2x-5 = y'[1-15y^2] \therefore
$$
$$
y' = \frac{2x-5}{1-15y^2}
$$
When at the point (4,1) you get
$$
y-1 = \frac{2(4)-5}{1-15}(x-4)
$$
$$
y = \frac{12-3x}{14}+1 = \frac{13}{7}-\frac{3x}{14}
$$
