By substituting for $b^2$, we have $a^2=c^2+ac+bc$, so $a^2-ac=c^2+bc$ and $a-c=\frac{c(b+c)}{a}$. Then \[a^2-c^2=\frac{bc(b+c)(a+c)}{ab} = \frac{a^2b^2}{ab} = ab,\]so $a^2=c^2+ab$. Now substituting for $a^2$ in $a^2=c^2+ac+bc$ gives $ab = ac+bc\implies \frac{1}{c}=\frac{1}{a}+\frac{1}{b}$.

Proposed by Vladimir Bragin

From the given equations we get $\frac{a^{2}}{b}=b+c$ and $\frac{b^2}{c}=c+a$, multiplying these and using results from the given forms we get $$\frac{a^2 b}{c}=(b+c)(c+a)=bc+ab+c^2 +ac=bc+ab+b^2 -ac+ac=bc+ab+b^2 =bc+ab+a^2 -bc=a(a+b) \Leftrightarrow \frac{ab}{c}=(a+b)$$(as $a,b,c \in \mathbb{R}^{+}$) $\Leftrightarrow \frac{1}{c}=\frac{a+b}{ab} = \frac{1}{a} + \frac{1}{b}$

Metropolises 2020 P4 wrote: Given three positive real numbers $a,b,c$ such that following holds $a^2=b^2+bc$, $b^2=c^2+ac$ Prove that $\frac{1}{c}=\frac{1}{a}+\frac{1}{b}$. Take $x=\frac{1}{a} y =\frac{1}{b}, z=\frac{1}{c}$, hence the given relations rewrite as $y^2z=x^2z+x^2y$ and $z^2x=y^2x+y^2z$. Hence, $$y^2z=x(z^2-y^2)=x(z-y)(z+y)=(xz+xy)(z-y)=\frac{y^2z}{x}(z-y) \Rightarrow x=z-y,$$that is, $z=x+y \Rightarrow \frac{1}{c}=\frac{1}{a}+\frac{1}{b}$ and we are done.

IOM P4 wrote: Given three positive real numbers $a,b,c$ such that following holds $a^2=b^2+bc$, $b^2=c^2+ac$ Prove that $\frac{1}{c}=\frac{1}{a}+\frac{1}{b}$. Rewrite the conditions as \begin{align} a^2 &= b(b+c) \\ b^2 &= c(c+a) \end{align}So, \begin{align*} && ca &= b^2 - c^2 && \text{(By (2))}\\ && &= (b+c)(b-c) \\ && &= \frac{a^2}{b}(b-c) && \text{(By (1))}\\ \implies && bc &= ab -ac \\ \implies && c(a+b) &= ab \\ \implies && \frac{1}{c} &= \frac{1}{a} +\frac{1}{b} \end{align*}

This is obvious due to a triangle of angles 1:2:4.

Since the equations are homogeneous, WLOG assume $c=1$, then $$a^2-a-1=b$$and $$(a^2-a-1)^2=1+a$$which implies $a^3-2a^2-a+1=0$ Hence $$ab-a-b=a^3-a^2-a-a^2+a+1-a=a^3-2a^2-a+1=0$$which implies $$\frac{1}{c}=1=\frac{1}{a}+\frac{1}{b}$$as desired.

Using the two equations, we derive \[\frac{a^2}b=b+c=\frac{ac}{b-c},\]from which the conclusion follows.

Since everything is homogenous, WLOG $c=1$. We have $a^2=b^2+b$ and $b^2=a+1$. Isolating $a$ in the second equation and squaring gives us that: $$b^4-2b^2+1=(b^2-1)^2=a^2=b^2+b\Rightarrow b^4-3b^2-b+1=0.$$Since $b\ne-1$, we can divide by $b+1$ to get $b^3-b^2-2b+1=0$. Now: $$\frac1a+\frac1b=\frac1{b^2-1}+\frac1b=\frac{b^2+b-1}{b^3-b}=\frac{b^2+b-1}{b^2+b-1}=1$$as desired.

Another proof: Construct an isosceles trapezoid $ABCD$ such that $AB=BC=CD=c$，$AC=BD=b$ (it's easy to verify that $b \geq 2c$ does not work)，since it must be cyclic, by Ptolemy's theorem, $AD=a$. Take the point $F$ such that $AF$ and $CD$ share a perpendicular bisector, then $ACDF$ is also an isosceles trapezoid. Then we have $AF=b$, again using Ptolemy's theorem. Since $AF=DF(=b)$, $F$ lies on the perpendicular bisector of $AD$, therefore we must also have $BF=CF(=a)$. Now that the five points $A, B, C, D, F$ lie on the same circle. Applying the Ptolemy's theorem a third time - this time on the quadrilateral $ABCF$. You don't even need to state that $ABCDEFG$ is a regular heptagon for some $E$ and $G$.