Optimization using KKT of a 3 variable function I want to maximize the function :
$$\sum_{i=0}^n x_i*ln(1+ \frac{ c*y_i*z_i }{x_i})$$ 
subject to :  $$\sum_{i=0}^n x_i \le X_0 \;\;\;\;\;\;\; and  \;\;\;\;\;\;\;  \sum_{i=0}^n y_i \le Y_0 $$
$$ x_i \;  and \;  y_i \; and \; z_i \; are \; Non-negative \;\;\;\;\;\;\;\;\;\;\;\;\;\; c \; and \; X_0 \; and \; Y_0 \; are \  positive \; constants $$
It is required to form the lagrangian, use KKT conditions and get $ x_i \; and \; y_i \; in \; terms \; of \; z_i $ if possible
I formed the lagrangian as follows : $ $
$$  \sum_{i=0}^n x_i*ln(1+ \frac{ c*y_i*z_i }{x_i}) +\lambda_1*(X_0 - \sum_{i=0}^n x_i) \; +\lambda_2*(Y_0 - \sum_{i=0}^n y_i) $$ 
I differentiated w.r.t.   y and equated the derivative to zero to get 
$$ x*\frac{ \frac{c*z}{x} }{ 1+\frac{c*y*z}{x} } -\lambda_2 = 0 $$
then w.r.t. z to get ( which seems trivial )
$$ x*\frac{ \frac{c*y}{x} }{ 1+\frac{c*y*z}{x} } = 0 $$
and finally w.r.t x to get 
$$ ln(1+ \frac{ c*y_i*z_i }{x_i}) + x*\frac{ \frac{-c*y*z}{x^2} }{1+\frac{c*y*z}{x}}     -\lambda_1 =0 $$
Am I right so far and shall continue ? as I am still a beginner. Thanks in advance
 A: Since you are only optimizing over $x=(x_1,x_2,\dots,x_n)$ and $y=(y_1,y_2,\dots,y_n)$ based on your above comment, the optimization problem has $2n$ variables. Writing it in minimization form,
\begin{align*}
&\text{minimize} && f(x,y) = -\sum_{i=1}^n x_i\log\left( 1 + \frac{cz_i y_i}{x_i} \right) \\
&\text{subject to} && \mathbf{1}^\top x \le X_0, \\
&&& \mathbf{1}^\top y \le Y_0, \\
&&& x \ge 0, ~ y\ge 0,
\end{align*}
where $c,X_0,Y_0\in\mathbb{R}$ and $z\in\mathbb{R}^n$ are given, and $\mathbf{1}$ denotes the $n$-vector of all ones. The Lagrangian for this problem is
\begin{equation*}
L(x,y,\lambda,\mu,\nu,\eta) = f(x,y) + \lambda(\mathbf{1}^\top x - X_0) + \mu(\mathbf{1}^\top y - Y_0) - \nu^\top x - \eta^\top y.
\end{equation*}
Now, the Lagrangian stationarity condition of KKT requires us to differentiate $L$ with respect to $x$ and $y$. To do so, note that
\begin{align*}
\frac{\partial f}{\partial x_j}(x,y) ={}& - \sum_{i=1}^n \left( \frac{\partial x_i}{\partial x_j}\log\left( 1+\frac{cz_i y_i}{x_i} \right) - x_i\frac{1}{1+\frac{cz_i y_i}{x_i}} \frac{cz_iy_i}{x_i^2}\frac{\partial x_i}{\partial x_j}\right) \\
={}& -\left( \log\left( 1+\frac{cz_jy_j}{x_j} \right) - \frac{1}{1+\frac{cz_jy_j}{x_j}}\frac{cz_jy_j}{x_j} \right),
\end{align*}
for $j\in\{1,2,\dots,n\}$. Taking a similar approach for differentiating with respect to $y$, we find that
\begin{equation*}
\frac{\partial f}{\partial y_j}(x,y) = -x_j\frac{1}{1+\frac{cz_jy_j}{x_j}}\frac{cz_j}{x_j} = - \frac{cz_j}{1+\frac{cz_jy_j}{x_j}}.
\end{equation*}
Now, putting these partial derivatives into vectors, we can write the gradient of $f$ in short form:
\begin{align*}
\nabla_x f(x,y) ={}& \left( \frac{\partial f}{\partial x_1}(x,y),\frac{\partial f}{\partial x_2}(x,y),\dots,\frac{\partial f}{\partial x_n}(x,y) \right), \\
\nabla_y f(x,y) ={}& \left( \frac{\partial f}{\partial y_1}(x,y),\frac{\partial f}{\partial y_2}(x,y),\dots,\frac{\partial f}{\partial y_n}(x,y) \right).
\end{align*}
With these gradients known, the Lagrangian stationarity condition becomes
\begin{align*}
\nabla_xL(x^*,y^*,\lambda,\mu,\nu,\eta) ={}& \nabla_x f(x^*,y^*) + \lambda\mathbf{1} - \nu = 0, \\
\nabla_yL(x^*,y^*,\lambda,\mu,\nu,\eta) ={}& \nabla_y f(x^*,y^*) + \mu\mathbf{1} - \eta = 0,
\end{align*}
which looks similar to your result aside from your missing Lagrange multiplier associated with the nonnegativity constraints and your indexing notation.
