Backpropagation with binary cross entropy loss formula I will classify using a neural network algorithm. I use 2 output, Y1=1 (positive) and Y2=0 (negative). The architecture is as follows:

loss that I use is binary cross entropy with the following formula:

From the loss, the error gradient with respect to weight v is:

based on the architecture that I demonstrated, is it correct to calculate the error gradient with respect to v11 like this?

 A: The substitution you made is correct. The formula you use is also correct assuming sigmoid activation at each layer:
According to the architecture, the forward and backward propagation equations are:
$$\begin{align}Z_j&=\sigma\left(\sum_{i=1}^{n_{input}} X_iV_{ij}+b_j\right)\rightarrow \frac{\partial Z_j}{\partial V_{ij}}=X_iZ_j(1-Z_j)
\\Y_k&=\sigma\left(\sum_{j=1}^{n_{hidden}} Z_jW_{jk}+b_k\right)\rightarrow \frac{\partial Y_k}{\partial Z_j}=Y_k(1-Y_k)W_{jk}
\\\frac{\partial Y_k}{\partial V_{ij}}&=\sum_{j'=1}^{n_{hidden}}\frac{\partial Y_k}{\partial Z_{j'}}\frac{\partial Z_{j'}}{\partial V_{ij}}=\frac{\partial Y_k}{\partial Z_{j}}\frac{\partial Z_j}{\partial V_{ij}}=Y_k(1-Y_k)W_{jk}X_iZ_j(1-Z_j)
\\E&=-\sum_{k=1}^{n_{out}}Q_k\log Y_k+(1-Q_{k})\log(1-Y_k)\rightarrow \frac{\partial E}{\partial Y_k}=\frac{Y_k-Q_k}{Y_k(1-Y_k)}\end{align}$$
Using the chain rule, we have
$$\begin{align}\frac{\partial E}{\partial V_{ij}}&=\sum_{k=1}^{n_{out}}\frac{\partial E}{\partial Y_k}\frac{\partial Y_k}{\partial V_{ij}}=\sum_{k=1}^{n_{out}}(Y_k-Q_k)W_{jk}X_iZ_j(1-Z_j)\end{align}$$
And, substituting $i=j=1$ here yields the following:
$$\begin{align}\frac{\partial E}{\partial V_{11}}&=\sum_{k=1}^2(Y_k-Q_k)W_{1k}X_1Z_1(1-Z_1)\\&=(Y_1-Q_1)W_{11}X_1Z_1(1-Z_1)+(Y_2-Q_2)W_{12}X_1Z_1(1-Z_1)\end{align}$$
