Please I think i don't understand this...
$u(x,y)=f(x^2+y^2,2y)$
$m=x^2+y^2$
$n=2y$
$\frac{du}{dx}=\frac{df}{dm}(2x)+\frac{df}{dn}0$
$\frac{du}{dy}=\frac{df}{dm}(2y)+\frac{df}{dn}2$
$\frac{d^2u}{dx^2}=(\frac{df}{dm}(\frac{df}{dm})(2x)+\frac{df}{dm}(\frac{df}{dn})(0))(2x)+\frac{df}{dm}(2)$
$\frac{d^2u}{dy^2}=(\frac{df}{dm}(\frac{df}{dm})(2y)+\frac{df}{dm}(\frac{df}{dn})(2))(2y)+\frac{df}{dm}(2) + (\frac{df}{dn}(\frac{df}{dm})(2y)+\frac{df}{dn}(\frac{df}{dn})(2))(2)+\frac{df}{dn}(0)$
Is this wrong? If it is, can u give me answer?
Edit: I don't care about $\frac{du}{dxdy}$, beacause i need to calcualte $\Delta u=\frac{d^2u}{dx^2}+\frac{d^2u}{dy^2}$
$$\frac{du}{dx}=\frac{d(x^2+y^2)}{dx}f_x+\frac{d(2y)}{dx}f_y=2xf_x$$
$$\frac{du}{dy}=\frac{d(x^2+y^2)}{dy}f_x+\frac{d(2y)}{dy}f_y=2yf_x+2f_y$$
$$\frac{d^2u}{dx^2}=\frac{d}{dx}\frac{du}{dx}=2f_x+2x(2xf_{xx})$$
$$\frac{d^2u}{dy^2}=\frac{d}{dy}\frac{du}{dy}=2f_x+2y(2yf_{xy}+2f_{yy})+2(2yf_{xy}+2f_{yy})$$