 Solution

Let $a=f(b,c)$ and work where both $f_b$ and $f_c$ are nonzero, so the required inverse partial derivatives exist. At fixed $a$, the total differential gives $0=f_{b}db+f_{c}dc$, hence $(\partial b/\partial c{)}_a=-f_c/f_b$. At fixed $b$, $(\partial c/\partial a{)}_b=1/f_c$. Their product with $f_b$ is $-1$: this is the cyclic partial derivative identity.

For explicit verification in this example, put $P=b^4-6b^2{c}^2+c^4$ and $H=(b^2+c^2{)}^3$. Direct differentiation givesThereforeThe identity is local to regular branches: for these explicit formulas require $bcP\ne 0$. At singular points one cannot justify cancellation of nonexistent inverse derivatives merely by writing a formal product.