Answer
$y\dfrac{\partial z}{\partial x}+x\dfrac{\partial z}{\partial y}=x$
Work Step by Step
Given: $z=y+f(x^2-y^2)$
Need to prove that $y\dfrac{\partial z}{\partial x}+x\dfrac{\partial z}{\partial y}=x$
Now, $\dfrac{\partial z}{\partial x}=2xf'(x^2-y^2)$
and $\dfrac{\partial z}{\partial y}=1-2yf'(x^2-y^2)$
Thus,
$y\dfrac{\partial z}{\partial x}+x\dfrac{\partial z}{\partial y}=x$
or, $y(2xf'(x^2-y^2))+x(1-2yf'(x^2-y^2))=x$
or $2xyf'(x^2-y^2)+x-2xyf'(x^2-y^2)+x=x$
or, $x=x$
Hence, it has been proved that $y\dfrac{\partial z}{\partial x}+x\dfrac{\partial z}{\partial y}=x$
