Calculate area of the surface $z = kxy/(1-x)$

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Calculate area of the surface: $$z = kxy/(1-x)$$ (https://www.geogebra.org/3d/uca6ehtc)

It's domain is restricted: $0\leq x\leq 1, 0\leq y\leq 1-x$

$k$ is just a parameter controlling its streching. I need a formula or maybe an integral to calculate the area of the surface.

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You can use the following formula to calculate the area of a surface: $$\iint_{S} dS$$ And you can find $dS$ as follow: $$dS = \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2} dA$$

So you end up with the following formula: $$ \iint_{S} dS = \iint_{D} \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2} dA $$

And for your problem it's: $$Area \space of \space z = \iint_{S} dS = \int_{0}^{1}\int_{0}^{1-x} \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2} dydx $$

You can read this also for more explanations: MIT18_02SC_MNotes_v9.3to4