Suppose \(z=f(x,y)\) is defined as an implicit differentiable function of \(x\) and \(y\), by an equation of the form \(F(x, y, z) = 0\), where \(F\) is also differentiable. If we use the chain rule to differentiate \(F\) with respect to \(x\):
$$ \frac{∂F}{∂x}\frac{∂x}{∂x} + \frac{∂F}{∂y}\frac{∂y}{∂x} + \frac{∂F}{∂z}\frac{∂z}{∂x} = 0$$
Since \(\frac{∂x}{∂x} = 1\), and \(\frac{∂y}{∂x} = 0\):
$$ \frac{∂F}{∂x} + \frac{∂F}{∂z}\frac{∂z}{∂x} = 0$$
If \(\frac{∂F}{∂z} \ne 0\), we can solve for \(\frac{∂z}{∂x}\) and obtain:
$$ \frac{∂z}{∂x} = -\frac{\frac{∂F}{∂x}}{\frac{∂F}{∂z}}$$
If we differentiate \(F\) with respect to \(y\), we can get \(\frac{∂z}{∂y}\) in a similar manner:
$$ \frac{∂z}{∂y} = -\frac{\frac{∂F}{∂y}}{\frac{∂F}{∂z}}$$