= 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 $P=b^4-6b^2c^2+c^4$ and $H=(b^2+c^2)^3$. Direct <differentiation> gives
$$
\left(\frac{\partial a}{\partial b}\right)_c=-\frac{cP}{H},\quad
\left(\frac{\partial a}{\partial c}\right)_b=\frac{bP}{H},\quad
\left(\frac{\partial b}{\partial c}\right)_a=\frac bc,\quad
\left(\frac{\partial c}{\partial a}\right)_b=\frac H{bP}.
$$
Therefore
$$
\boxed{\left(\frac{\partial a}{\partial b}\right)_c\left(\frac{\partial b}{\partial c}\right)_a\left(\frac{\partial c}{\partial a}\right)_b=-1.}
$$
The identity is local to regular branches: for these explicit formulas require $bcP\ne0$. At singular points one cannot justify cancellation of nonexistent inverse derivatives merely by writing a formal product.
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