Calculating the tangent in line given an implicitly given curve. Given the implicitly given curve $x^4y^2-2\frac{y}{x}=ln(y-1)$ and the point P = (-1, 0).

Calculate the tangent in P to the curve.

I've tried to calculate the partial derivatives for x and y, and then using the $z - z_0 = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)$ to get the tangent line, but what's $z$? 

Or am I trying to solve this in the wrong way? 

The answer should be $x + y + 1 = 0$

Here's the original problem (in Dutch).
$$x^4y^2-2\frac{y}{x}=ln(y-x)\\f(x,y)= x^4y^2-2\frac{y}{x}-ln(y-x)=0\\y'=-\frac{4x^3y^2-2\frac{-y}{x^2}-\frac{-1}{y-x}}{2x^4y-2\frac{1}{x}-\frac{+1}{y-x}}$$
$$x^4y^2-2\frac{y}{x}=ln(y-x)\\f(x,y)= x^4y^2-2\frac{y}{x}-ln(y-x)=0\\y'=-\frac{4x^3y^2-2\frac{-y}{x^2}-\frac{-1}{y-x}}{2x^4y-2\frac{1}{x}-\frac{+1}{y-x}}$$