The criterion for exactness says that, given $M,N \in C^1(D)$, a differential equation $Mdx+Ndy=0$ is exact if and only if $M_y=N_x$.
We know that
$$M_y =1 + x \frac{\partial}{\partial y} f(x^2+y^2) \\ N_x = -1 + y \frac{\partial}{\partial x} f(x^2+y^2)$$
Hence, if $M_y=N_x$,
$$2 = y\frac{\partial}{\partial x} f(x^2+y^2) – x \frac{\partial}{\partial y} f(x^2+y^2)$$
But I don't how to proceed (or how to interpret this identity, really). If $f$ is a one-variable function, do these partials mean anything?
Best Answer
$$M_y =1 + x \frac{\partial}{\partial y} f(x^2+y^2)$$ $$N_x = -1 + y \frac{\partial}{\partial x} f(x^2+y^2)$$ Let $X(x,y)=x^2+y^2$
$\frac{\partial}{\partial x} f(x^2+y^2)=\frac{\partial}{\partial x} f(X)=f'(X)\frac{\partial X}{\partial x}=2xf'(X)$
$\frac{\partial}{\partial y} f(x^2+y^2)=\frac{\partial}{\partial y} f(X)=f'(X)\frac{\partial X}{\partial y}=2yf'(X)$
$$M_y =1 + x \frac{\partial}{\partial y} f(x^2+y^2)=1+x\big(2yf'(X) \big) = 1+2xyf'(X)$$ $$N_x = -1 + y \frac{\partial}{\partial x} f(x^2+y^2)=-1+y\big(2xf'(X) \big) =-1+2xyf'(X)$$ Since $1+2xyf'(X)\neq -1+2xyf'(X)\quad\implies\quad M_y\neq N_x$